Ask question

Ask question

All categories

murzikaleks [220]

3 years ago

9

Part A: Virginia earned $14 walking her neighbors' dogs on Saturday. She earned some extra money on Sunday doing the same thing.

Write an expression with a variable that shows the total amount of money Virginia earned Saturday and Sunday.
Part B: Virginia was able to walk 2 less than twice as many dogs as her friend Alicia. Write an algebraic expression to represent the number of dogs Virginia walked compared with Alicia.

1 answer:

shutvik [7]3 years ago

7 0

Part A: 14 + 14 = y or 14•2 = y
Part B: (n•2) - 2 = a

Not 100% sure that this is correct but, hope this helps! :) If you have any questions than ask me.

You might be interested in

A rectangle is 6yards longer than it is wide. Find the dimensions of the rectangle if it’s area is 187 square yards

emmasim [6.3K]

Answer:

Width = 11 yards

Length = 17 yards

Step-by-step explanation:

First of all, the length of the rectangle is 6 yards longer than the width, this means, length = width + 6 yards. This dimensions can be represented on figure 1, where <em>w</em> is width, and <em>l</em>, for length.

We know the area of a rectangle is A = width x length

For our case 187 = w . (w + 6)

Using the Distributive Property for the multiplication we obtain

$187 = w^{2} +6w$

$w^{2} +6w-187 =0,$

Using the quadratic formula $w=\frac{-b\±\sqrt{b^{2}-4ac } }{2a}$ where a = 1, b = 6, c = - 187 and replacing into the formula, we will have:

$w=\frac{-6\±\sqrt{6^2-4(1)(-187)} }{2(1)}$

$w=\frac{-6\±\sqrt{36+748} }{2}=\frac{-6\±\sqrt{784} }{2}=\frac{-6\±28}{2}$

We have two options: $w=\frac{-6+28}{2}=\frac{22}{2}=11 yards$

Or

$w=\frac{-6-28}{2}=\frac{-34}{2}=-17 yards$ But a distance (width) can not be negative so, this answer for w must be discarded.

The answer must be width = 11 yards.

To find the length $l =\frac{187}{11}=17 yards$

6 0

3 years ago

Write the equation in slope-intercept form. -10x+2y=12<br><br> -please help

Goryan [66]

Answer:

y=5x+6

hope this helps

have a good day :)

Step-by-step explanation:

3 0

3 years ago

Show all your work to find the axis of symmetry of f(x) = −2(x − 4)^2 + 2.

Naya [18.7K]

Answer:

-2(x^2+15)

Step-by-step explanation:

f(x) = −2(x − 4)^2 + 2.

Expanding (x-4)^2

(x-4)(x+4)

x(x+4)-4(x+4)

x^2+4x-4x-16

x^2-16

F(x) = -2(x^2-16)+2

-2x^2-32+2

-2x^2-30

-2(x^2+15)

6 0

3 years ago

Which algebraic expression means “four less than three times a number”?

Monica [59]

It’s 3n-4

I might be right or wrong

5 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

4 years ago