3 years ago

What is 140 rounded to the nearest tenth of a percent

Mathematics

1 answer:

Ierofanga [76]3 years ago

3 0

140 changed to a percent would be 1.4 (divide by 100), and rounded the nearest tenth would be 1.4.

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Given that (-3,7) is on the graph of f(x), find the corresponding point for the function f(x + 5).

Stella [2.4K]

We have that
(-3,7) is on the graph of f(x), find the corresponding point for the function
f(x + 5)

we know that
in the transformation f(x)-------> f(x+5)
the point (x.y) in f(x) becomes a point (x-5,y) in f(x+5)
so
(-3,7)-----------> becomes a point (-3-5,7)------> (-8,7)

the answer is
(-8,7)

7 0

3 years ago

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Tamiku [17]

Answer:

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4 years ago

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Leona [35]

Answer:

X=-5y/13+90/13, Y=-13x/5+18

Step-by-step explanation:

You need to solve the equation for X and Y.

Solving for X:

13x+5y=90

Subtract 5y: 13x=-5y+90

Divide by 13: x=-5y/13+90/13

For Y:

13x+5y=90

Subtract 13x: 5y=-13x+90

Divide by 5: y=-13x/5+90/5

(Simplifies to -13x/5+18)

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3 years ago

Solve for the following 3z-5+2z+25-5z

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Z+30 is your answer..

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4 years ago

Someone solve this for me please: Find a vector v such that || v || = 10 and v is in same direction as 2i + 3j

drek231 [11]

Answer:

Step-by-step explanation:

$\overrightarrow{a}=2*\overrightarrow{i}+3*\overrightarrow{j}\\\\||\overrightarrow{a}||=\sqrt{2^2+3^2} =\sqrt{13} \\\\\boxed{\overrightarrow{v}=\dfrac{20}{\sqrt{13}}*\overrightarrow{i}+\dfrac{30}{\sqrt{13}}*\overrightarrow{j}} \\\\\\||\overrightarrow{a}||=\sqrt{(\dfrac{20}{\sqrt{13}})^2+(\dfrac{30}{\sqrt{13}})^2} =\sqrt{\dfrac{400+900}{13} }==\sqrt{\dfrac{1300}{13} }=\sqrt{100}=10 \\$

8 0

3 years ago