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3 years ago

Write an equation in slope intercept form and in standard form for the line passing through (4,9) and perpendicular to x=-6

Mathematics

1 answer:

Ivahew [28]3 years ago

4 0

Answer:

Input work is the work done on a machine as the input force acts through the input distance. This is in contrast to output work which is a force that is applied by the body or system to something else. Output work is the work done by a machine as the output force acts through the output distance.Jan 15, 2020 hwgsvsd d jxvsd

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How to simplify a product

Stella [2.4K]

Answer:

simple marketing

Step-by-step explanation:

if you have simple marketing that is easily understandable then it will most likely get better sales

8 0

3 years ago

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Write an expression by subtracting one expression in terms of x from the other

Eva8 [605]

I believe the answer is 128 because span span

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3 years ago

ABCD is a rectangle. Find the measure of the angle CWD

Kruka [31]

Answer:

120 degrees.

Step-by-step explanation:

Angles in a triangle add up to 180 degrees and triangle BWA is an isosceles triangle, meaning that AW and BW are congruent, or equal in length to each other.

This also means that angle WBA and WAB are congruent to each other. Angle WAB is 30 degrees, so angle WBA is also 30 degrees.

Knowing that angles in a triangle add up to 180 degrees we have 30+30+BWA=180.

60+BWA=180.

BWA=120 degrees.

Hope this helps!

6 0

4 years ago

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Which expressions are equivalent to 42-5? Choose all the correct answers. Show your work in arriving at your answers.

frozen [14]

Answer:

B, C, E

Step-by-step explanation:

<u />

<u>Exponent Rules</u>

$a^b \cdot a^c=a^{b+c}$

$\dfrac{a^b}{a^c}=a^{b-c}$

$a^n \cdot b^n=(ab)^n$

$a^0=1$

$\dfrac{1}{a^b}=a^{-b}$

$(a^b)^c=a^{bc}$

$\textsf{A}\quad42^{-3} \cdot 42^8=42^{-3+8}=42^5$

$\textsf{B}\quad \dfrac{42^{-7}}{42^{-2}}=42^{-7-(-2)}=42^{-5}$

$\textsf{C}\quad \dfrac{6^{-5}}{7^5}=6^{-5} \cdot 7^{-5}=(6 \cdot 7)^{-5}=42^{-5}$

$\textsf{D}\quad 42^0 \cdot 42^5=1 \cdot 42^5=42^5$

$\textsf{E}\quad \left(\dfrac{1}{42^{-5}}\right)^{-1}=(42^5)^{-1}=42^{-5}$

$\textsf{F}\quad \left(\dfrac{6^{-5}}{7^5}\right)^{-1}=\left(6^{-5} \cdot 7^{-5}\right)^{-1}=\left(\left(6 \cdot 7\right)^{-5}\right)^{-1}=42^{-5(-1)}=42^5$