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3 years ago

15

Tanvi teaches art lessons at a day camp and needs to get supplies. She purchases 40 Bright Streak paint sets from the Color a sp

lash art store. She uses a coupon she received in the mail for $10 off her purchase. Tanvi spends $310 in all

1 answer:

natima [27]3 years ago

4 0

Answer:

The paint sets cost $8 each.

Step-by-step explanation:

Tanvi pays $310 after getting a $10 discount. Before the disscount her total was $320. 320 divided by 40 sets is $8 per set.

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I need help pls and thank u

spayn [35]

Find the difference between the midpoint and the known endpoint by subtracting the X and Y values from each other:

X: 10 - 6 = 4

Y: 7 - 4 = 3

Now add those values to the midpoint to find the other endpoint:

X: 10 +4 = 14

Y: 7 +3 = 10

Endpoint = (14,10)

4 0

3 years ago

Can someone help me please

Marrrta [24]

Answer:

y = 3x - 6

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, - 6) and (x₂, y₂ ) = (2, 0) ← 2 points on the line

m = $\frac{0-(-6)}{2-0}$ = $\frac{0+6}{2}$ = $\frac{6}{2}$ = 3

the line crosses the y- axis at (0, - 6 ) ⇒ c = - 6

y = 3x - 6 ← equation of line L

3 0

2 years ago

Midpoint Between Two Given Points

kondaur [170]

Answer:

(1, -7)

Step-by-step explanation:

the difference between -4 & 6 is 10 so you add 5 (half) to -4 or subtract from 6, this gives you 1

you do the same for -9 and -5, the midpoint for those being -7, I hope this helps

4 0

3 years ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

Solve please?<br><br>xd=y(L-d)<br><br>please explain the steps also

RSB [31]

Xd = y(L-d) then divide by "y"
xd/y = L - d then add "d"
xd/y + d = L

4 0

4 years ago

Read 2 more answers