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3 years ago

The function fx) =ax-4 will never cross the x-axis if a is positive. A. True B. False

Mathematics

1 answer:

boyakko [2]3 years ago

8 0

Answer:

False because a linear equation always crosses the x axis (unless x is zero, which it isn't in this case).

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What is the answer and method to find the quotient of 5.25/1.5

Mademuasel [1]

Answer: 3.5

Use long division

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3 years ago

For questions 4-5, find the volume of the cylinder in terms of pi.

dangina [55]

Answer:

hope it helps you!!!!!!!!

6 0

3 years ago

2(x +5)=(x-4) ?<br> Step by step

ExtremeBDS [4]

$\text {Hi There! Let's Solve this Problem!}$

$\text {\underline {The First Step is to Simplify}}$

$\text {2*x=2x}\\\text {2*5=10}$

$\text {Your New Problem Is: 2x+10=x-4}$

$\text {\underline {The Next Step is to Subtract x}}$

$\text {2x+10-x=x-4-x}$

$\text {Your New Problem Is: x+10=-4}$

$\text {\underline {The Final Step is to Subtract 10}}$

$\text {x+10-10 (Crosses Out)}\\\text {-4-10}$

$\text {Your Answer Would Be:}$

$\Huge\boxed {x=-14}$

$\text {Best of Luck!}$

7 0

3 years ago

Read 2 more answers

Abed says he has written a system of two linear equations that has an infinite number of solutions. One of the equations of the

LenKa [72]

Answer:

3x - y = 1

Step-by-step explanation:

This will be the same equation written in a different way'

The original equation is:

y = 3x - 1

0 = 3x - y - 1

3x - y = 1.

3 0

3 years ago

In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find: The length of the angle bisector of a

Aliun [14]

Check the picture below.

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\quad cos(CAB)=\cfrac{5}{13}\implies \measuredangle CAB=cos^{-1}\left( \frac{5}{13} \right)
\\\\\\
\textit{that means that }\measuredangle hAC=\cfrac{cos^{-1}\left( \frac{5}{13} \right)}{2}\impliedby \textit{one of the halves}$

now, notice, for the angle hAC, the hypotenuse is hA, and the adjacent side is CA, therefore,

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(hAC)=\cfrac{5}{hA}\implies hA=\cfrac{5}{cos(hAC)}
\\\\\\
hA=\cfrac{5}{cos\left[ \frac{cos^{-1}\left( \frac{5}{13} \right)}{2} \right]}$

make sure your calculator is in Degree mode, if you need the angle in degrees.

8 0

3 years ago