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3 years ago

A carpenter charges a flat fee of $150 plus $45 per hour. Which table represents this situation?

Mathematics

2 answers:

Inessa05 [86]3 years ago

8 0

Answer is C.

A carpenter charges a flat fee of $150 plus $45 per hour.

Total = 150 + 45h

Hope it helps

mezya [45]3 years ago

7 0

So the first hour is 195 dollars. so we take 195 and multiply it by 2 to get 390. There is only one table that has 195 for the first and 390 btu we are gnna keep going. So next is 585, And finally 780. Your answer is A.

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How many solutions can be found for the set of linear equations? Y = 7(x + 2) and y = 7x + 14

Bogdan [553]

Answer:

infinite solution

Step-by-step explanation:

because Y = 7(x + 2) and y = 7x + 14 are the same

y = 7(x+2) = 7x + 14

eliminate to see

y = 7x + 14

-(y = 7x +14)

= 0 = 0 + 0

so it is infinite solution

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3 years ago

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Choose Yes or No to tell whether each value of p is a solution for this inequality.

blsea [12.9K]

No
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3 years ago

A candy company can make 360 pounds of chocolate in 6 hours how much can they make in 10 hours

LiRa [457]

Answer:

600

Step-by-step explanation:

divide 360 by 6 and you get how many pounds that can make in one hour.

which is 60 so then you multiply 60 by 10 and get 600

3 0

3 years ago

Find the equation of the line tangent to the function at the given point f(x)=5x^2 at x=10

Svet_ta [14]

f(x) = 5x²

f(10) = 5(10)² = 5(100) = 500

f'(x) = 10x

f'(10) = 10(10) = 100

Now, find the line that passes through (10, 500) and has a slope of 100

y - y₁ = m(x - x₁)

y - 500 = 100(x - 10)

y - 500 = 100x - 1000

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8 0

3 years ago

Can someone please help me on number 16-ABC

melomori [17]

Answer:

Please check the explanation.

Step-by-step explanation:

Given the inequality

-2x < 10

-6 < -2x

Part a) Is x = 0 a solution to both inequalities

FOR -2x < 10

substituting x = 0 in -2x < 10

-2x < 10

-3(0) < 10

0 < 10

TRUE!

Thus, x = 0 satisfies the inequality -2x < 10.

∴ x = 0 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 0 in -6 < -2x

-6 < -2x

-6 < -2(0)

-6 < 0

TRUE!

Thus, x = 0 satisfies the inequality -6 < -2x

∴ x = 0 is the solution to the inequality -6 < -2x

Conclusion:

x = 0 is a solution to both inequalites.

Part b) Is x = 4 a solution to both inequalities

FOR -2x < 10

substituting x = 4 in -2x < 10

-2x < 10

-3(4) < 10

-12 < 10

TRUE!

Thus, x = 4 satisfies the inequality -2x < 10.

∴ x = 4 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 4 in -6 < -2x

-6 < -2x

-6 < -2(4)

-6 < -8

FALSE!

Thus, x = 4 does not satisfiy the inequality -6 < -2x

∴ x = 4 is the NOT a solution to the inequality -6 < -2x.

Conclusion:

x = 4 is NOT a solution to both inequalites.

Part c) Find another value of x that is a solution to both inequalities.

solving -2x < 10

$-2x\:$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)>10\left(-1\right)$

Simplify

$2x>-10$

Divide both sides by 2

$\frac{2x}{2}>\frac{-10}{2}$

$x>-5$

$-2x-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-5,\:\infty \:\right)\end{bmatrix}$

solving -6 < -2x

-6 < -2x

switch sides

$-2x>-6$

Multiply both sides by -1 (reverses the inequality)

$\left(-2x\right)\left(-1\right)$

Simplify

$2x$

Divide both sides by 2

$\frac{2x}{2}$

$x$

$-6$

Thus, the two intervals:

$\left(-\infty \:,\:3\right)$

$\left(-5,\:\infty \:\right)$

The intersection of these two intervals would be the solution to both inequalities.

$\left(-\infty \:,\:3\right)$ and $\left(-5,\:\infty \:\right)$

As x = 1 is included in both intervals.

so x = 1 would be another solution common to both inequalities.

<h3>SUBSTITUTING x = 1</h3>

FOR -2x < 10

substituting x = 1 in -2x < 10

-2x < 10

-3(1) < 10

-3 < 10

TRUE!

Thus, x = 1 satisfies the inequality -2x < 10.

∴ x = 1 is the solution to the inequality -2x < 10.

FOR -6 < -2x

substituting x = 1 in -6 < -2x

-6 < -2x

-6 < -2(1)

-6 < -2

TRUE!

Thus, x = 1 satisfies the inequality -6 < -2x

∴ x = 1 is the solution to the inequality -6 < -2x.

Conclusion:

x = 1 is a solution common to both inequalites.

7 0

3 years ago