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3 years ago

This histogram shows the numbers of shoppers in various age groups at a clothing store.

Mathematics

2 answers:

Bond [772]3 years ago

7 0

Answer:

is it 7???

Step-by-step explanation:

alisha [4.7K]3 years ago

4 0

Th answer is b 7 I just took the test

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The table below represents an exponential function what is the common ratio

Greeley [361]

Solution:

Given:

A table representing an exponential function.

The x-values represent the terms, while the y-values represent the numbers.

The common ratio is gotten from the numbers (y-values).

The formula for common ratio is given by;

$\begin{gathered} r=\frac{present\text{ term}}{preceed\text{ ing term}} \\ \\ \text{Hence,} \\ r=\frac{108}{36}=\frac{36}{12}=\frac{12}{4}=\frac{4}{1.33\ldots}=\frac{1.33\ldots.}{0.44\ldots} \\ r=3 \end{gathered}$

Therefore, the common ratio is 3.

3 0

1 year ago

Sue deposited $52 into her bank account, so she currently has $12 in her account. What was her balance before the deposit?

Semenov [28]

Answer:

-40

Step-by-step explanation:

12 - 52 = -40

hope this helps...

7 0

3 years ago

Read 2 more answers

Find the components of the vertical force Bold Upper Fequalsleft angle 0 comma negative 8 right anglein the directions parallel

nydimaria [60]

Answer with Step-by-step explanation:

We are given that

F=<0,-8>=0i-8j=-8j

$\theta=\frac{\pi}{3}$

The component of force is divided into two direction

1.Along the plane

2.Perpendicular to the plane

1.The vector parallel to the plane will be= $r=cos\frac{\pi}{3}i-sin\frac{\pi}{3}j=\frac{1}{2}i-\frac{\sqrt 3}{2}j$

By using $cos\frac{\pi}{3}=\frac{1}{2},sin\frac{\pi}{3}=\frac{\sqrt 3}{2}$

Force along the plane will be= $\mid F_x\mid=F\cdot r$

Force along the plane will be = $\mid F_x\mid=F\cdot (\frac{1}{2}i-\frac{\sqrt 3}{2}j)=-8j\cdot(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=8\times \frac{\sqrt 3}{2}=4\sqrt 3$ N

By using $i\cdot i=j\cdoty j=k\cdot k=1,i\cdot j=j\cdot k=k\cdot i=j\cdot i=k\cdot j=i\cdot k=0$

Therefore, force along the plane= $\mid F_x\mid(\frac{1}{2}i-\frac{\sqrt 3}{2}j)=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)$

2.The vector perpendicular to the plane= $r=-sin\frac{\pi}{3}-cos\frac{\pi}{3}=-\frac{\sqrt 3}{2}i-\frac{1}{2}j$

The force perpendicular to the plane= $\mid F_y\mid=F\cdot r=-8j(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

The force perpendicular to the plane= $4$ N

Therefore, $F_y=4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $F_x+F_y=4\sqrt 3(\frac{1}{2}i-\frac{\sqrt 3}{2}j)+4(-\frac{\sqrt 3}{2}i-\frac{1}{2}j)$

Sum of two component of force= $2\sqrt 3i-6j-2\sqrt3 i-2j=-8j$

Hence,sum of two component of forces=Total force.

6 0

3 years ago

Help me please, needed asap

murzikaleks [220]

Answer:

4 fish and 3 cportion of chips- $28.35

Step-by-step explanation:

5 0

3 years ago

Find the domain of the function.

timurjin [86]

When taking square roots, you can't take square roots of negative roots of negative numbers. So, what will work for the domain of u(x) is what makes u(x) zero or more. We can make an inequality for that.

u(x) ≥ 0.

$\sqrt{9x+27} \geq 0$

9x + 27 ≥ 0 by squaring both sides

9x ≥ -27

x ≥ -3

So the domain of the function is when x ≥ -3 is true.

5 0

4 years ago