PLS HELP THIS IS DUE TODAY

Answer:

3 chocolate bars, and 2 graham crackers

Someone forgot the marshmallows...... :P

Step-by-step explanation:

Chocolate bars = 8 pack

Graham Crackers = 12pack

To have no crackers or chocolate left over, we need to find LCM

Factors of 8:

8, 16, 24, 32, 40, 48, 56, 72....

Factors of 12:

12, 24, 36, 48, 60, 72

The smallest LCM is 24

Chocolate bars:

24/8 = 3

Graham Crackers:

24/12 = 2

If my answer is incorrect, pls correct me!

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-Chetan K

Answer:

3 packs of chocolate and 2 packs of crackers

Step-by-step explanation:

the lowest common multiple of 8 and 12 is 24. We can determine this by prime factorization:

Prime factors of 8: 2 x 2 x 2

Prime factors of 12: 2 x 2 x 3

multipyling the bottom rungs of our factor tree we get: 2 x 2 x 2 x 3 = 24.

If you need me to draw the factor tree, just ask.

Answer:

12x^7.

Step-by-step explanation:

3x^3(4x^4)=

3*4 * x^(3+4)

= 12x^7.

Answer:

x = 65 degrees

Step-by-step explanation:

x° + x° + 50° = 180°

2x + 50° = 180°

2x = 180° - 50°

2x = 130°

x = 130°/2

x = 65°

Answer:

x/4+10=2

x/4=2-10

x/4=-8

x=-8 x 4

x=-32

Step-by-step explanation:

Answer:

equal or balanced

Step-by-step explanation

the = demonstrates the amounts need to be the same

Answer: Balanced.

Step-by-step explanation:

For an equation each side shall be equal (balanced) or else it's NOT an equation.

Answer:

6x+7y=56

Step-by-step explanation:

In order to have an equation that is parallel to another the slopes multiplied by one another must equal -1. In this case, the first slope is [tex]\frac{7}{6}[/tex] . So [tex]\frac{7}{6}[/tex]×what = -1

Normally you can just flip the numerator and denominator and add the opposite sign. So we can do that here. [tex]-\frac{6}{7} *\frac{7}{6} =-1[/tex]. So the equation should have [tex]-\frac{6}{7}[/tex].

But none of the answers obviously have that slope. If you try the second one, which looks different from the rest and will give you a negative x when you subtract it from both sides. Then you can divide by 7, you get [tex]y=-\frac{6}{7}x +8[/tex].

Answer:

28x-16y+45.4

Step-by-step explanation:

Add up all the variables inside the bracket by first grouping like terms

Then, multiply through by 2 ✅✅

Answer:

Does the answer help you?

Answer:

Ben needs to get a 99% on his second test

Step-by-step explanation:

The average is the sum of both test scores, divided by the total number of test scores (2).

Let x by the grade Ben must receive on his second test.

(89 + x)/2 = 94

Multiply by 2 on both sides

89 + x = 188

Subtract 89 on both sides

x = 99

Check work:

(89 + 99)/2 = 188/2 = 99

Answer: 99

Concept:

Here, we need to know the idea of the average value.

In mathematics, average refers to the sum of a group of values divided by n, where n is the number of values in the group.

Average = Sum of value / number of values

Solve:

Given information

Scoring on first test = 89

Scoring on second test = x

Number of tests = 2

Average = 94

Given expression

Average = Sum of value / number of values

Substitute values into the expression

94 = (89 + x) / 2

Multiply 2 on both sides

94 × 2 = (89 + x) / 2 × 2

188 = 89 + x

Subtract 89 on both sides

188 - 89 = 89 + x - 89

[tex]\boxed{x=99}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

[tex]28\sqrt{3}[/tex]

Step-by-step explanation:

The area of the big triangle is 1/2 b h =  1/2*6*(12^2 = 6^2 + x^2)

that ends up being [tex]\sqrt{108} = 36\sqrt{3}[/tex]

the small triangle are needs to be subtracted....

[tex]\frac{\left(4\cdot \:sin\left(90\right)\right)}{sin\left(30\right)}[/tex] that is the length of the unknown side...

1/2 B * h of that triangle get you to [tex]8\sqrt{3}[/tex]

just subtract the two areas

Answer:

(B) 28√3

Step-by-step explanation:

The area of quadrilateral ABED is equal to the area of triangle CDE subtracted from the area of triangle ABC.

Area of triangle CDE:

Triangle ABC is equilateral. All sides have length 12.

AB = BC = AC = 12

BE = 8

BE + EC = BC

8 + EC = 12

EC = 4

In an equilateral triangle, all angles measure 60°.

m<C = 60°

m<CDE = 30°

Triangle CDE is a 30-60-90 triangle.

DE = EC√3

DE = 4√3

area of triangle CDE = bh/2

area of triangle CDE = (EC)(DE)/2

area of triangle CDE = (4)(4√3)/2

area of triangle CDE = 8√3

Area of triangle ABC:

Side AC is a base of triangle ABC.

AC = 12

(1/2)AC = 6

The altitude of triangle ABC from side AC to vertex B measures

h = 6√3

area of triangle ABC = bh/2

area of triangle ABC = (AC)(h)/2

area of triangle ABC = (12)(6√3)/2

area of triangle ABC = 36√3

area of quadrilateral ABED = area of triangle ABC - area of triangle CDE

area of quadrilateral ABED = 36√3 - 8√3

area of quadrilateral ABED = 28√3

Answer:

z^2-3z-18

=(z^2+3z)+(-6z-18)

=z(z+3)-6(z+3)

=(z+3)(z-6)

Answer:

[tex]\\ \sf\longmapsto -5+(5+3)[/tex].

[tex]\\ \sf\longmapsto -5+5+3[/tex]

[tex]\\ \sf\longmapsto (-5+5)+3[/tex]

[tex]\\ \sf\longmapsto 0+3[/tex]

[tex]\\ \sf\longmapsto 3[/tex]

Answer:

1

Step-by-step explanation:

(1-x)/x=0

or, 1-x=0

or, -x=-1

or, x=1

Answer:

F(x) = 0 when x=1

Step-by-step explanation:

We need the numerator to equal zero

1-x = 0

Add x to each side

1-x+x = 0+x

1 =x

The denominator cannot equal zero( The denominator does not equal zero when x =1)

F(x) = 0 when x=1

Answer: [tex]\frac{\sqrt{5}}{6}[/tex]

==================================================

Work Shown:

[tex]\frac{\sqrt{10}}{3\sqrt{8}}\\\\\frac{\sqrt{10}*\sqrt{8}}{3\sqrt{8}*\sqrt{8}}\\\\\frac{\sqrt{10*8}}{3\sqrt{8*8}}\\\\\frac{\sqrt{80}}{3\sqrt{64}}\\\\\frac{\sqrt{80}}{3*8}\\\\\frac{\sqrt{16*5}}{24}\\\\\frac{\sqrt{16}*\sqrt{5}}{24}\\\\\frac{4*\sqrt{5}}{24}\\\\\frac{\sqrt{5}}{6}\\\\[/tex]

Note: In the second step, I multiplied top and bottom by [tex]\sqrt{8}[/tex] to rationalize the denominator.

Answer:

square root of 5 over 6 or 0.37267799…

Step-by-step explanation:

Multiply the numerator and denominator by the conjugate.

Answer:

4360

Step-by-step explanation:

MXC is 1090 and a IV is 4 so 1090 x 4 is 4360

Answer:

[tex] \sqrt{ {x}^{10} } \div x \\ \frac{ \sqrt{ {x}^{9}x } }{x} \\ = \sqrt{ {x}^{9} } \\ = \sqrt{ {x}^{8}x } \\ = {x}^{4} \sqrt{x} \\ thank \: you[/tex]

Answer:

x⁴√x

Step-by-step explanation:

simplify x¹⁰ and x to √x⁹ then simplify the radical equation to x⁴√x

Answer:

B 2 m

Step-by-step explanation:

2 cm and 2 mm are way to small they are less than the size of a cell phone

2 km is way to large it is larger than a car

Answer:

y(x)=10.5/sqrt(x)

Step-by-step explanation:

Since y is inversely proportional to the square root of x, it can be written as y=a/sqrt(x). To compute a, we will use the point given in the question (2.25,7). 7=a/sqrt(2.25), a=10.5. The function is y(x)=10.5/sqrt(x)

Answer:

x-12

Step-by-step explanation:

Let the variable be x

12 less than

x-12

2:3

Given Racheal = 24 years

Thomson = 2/3×24

= 2×8

= 16 years

Must click thanks and mark brainliest

Answer:

Step-by-step explanation:

This is a consistent dependent system. This means that the lines will eventually intersect in one place.

Inconsistent systems are parallel lines and equivalent systems represent a line completely overlapped by another line because they have the same equation.

The measure of SR is 10.

The given figure has two secant segments and two external segments.

QU = secant segment

QS = secant segment

QP = external segment of QU

QR = external segment of QS

We need to find the measure of SR.

It states that the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.

Applying the secant theorem in the given figure.

QU x QP = QS x QR

( 9 + 7 ) x 9 = (8 + 3X-5) x 8

16 x 9 =  (3 + 3X) x 8

144 = 24 + 24X

144 - 24 = 24X

120 = 24X

X = 120 / 24

X = 5

Now Putting X = 5 in SR = 3X-5

We get,

SR = 3 x 5 - 5

SR = 15 - 5

SR = 10

After applying the secant segment theorem we got the measure of SR as 10.

Learn more about secant segments in a circle here:

https://brainly.com/question/18760361

#SPJ5

Answer:

What is the vertical change from Point A to Point B? 1

What is the horizontal change from Point A to Point B? 2

What is the rate of change shown on the graph? Give

the answer as a decimal rounded to the nearest tenth.

if necessary: 0.5

Step-by-step explanation:

The rate of change, as the own name suggest, is the rate at which the function change its value. The function, in this case is changing with a rate of 0.5 because to each unit that x grows, y grows the half part of it.

Answer:

x= 0, and x=4

Step-by-step explanation:

x² -x  = 3x, we can subtract 3x from both sides

x² -x -3x = 0, combine like terms

x² -4x = 0 , factor x

x *( x-4) = 0, if the product of 2 numbers equals 0 at least one is 0

x=0 and

x-4 = 0, add 4 to both sides

x= 4

Answer:

x=4  simple thing with all problems like this, is if it has a exponent, then half the exponent and add it to the numerical digit on the right of the equals sign, and you have you answer

Step-by-step explanation:

Answer: 20 times 2 equal 40/ 40 divided by 2 is 20

Step-by-step explanation: Both of them should help you find the answer

Answer:

fohohcoufohohcouvhop

Step-by-step explanation:

typing mistake sorry

[tex]\\ \sf\longmapsto x+73=90[/tex]

[tex]\\ \sf\longmapsto x=90-73[/tex]

[tex]\\ \sf\longmapsto x=17[/tex]

Why?

Sum of two complementary angles is 90°

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

Simplify,

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

[tex]2-\sqrt{3}=\frac{(\sqrt{3}-1)^2}{3-1}[/tex]

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer:

10.49

Step-by-step explanation:

FG= squareroot(10)(11) =10.49 took quiz too

Answer:

x=4y

Step-by-step explanation:

3^x=3^4y

x=4y

am I correct?