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3 years ago

What is the range of the relation?

Mathematics

1 answer:

4vir4ik [10]3 years ago

6 0

Hello!

The range is all of the y-values in a set of points. They are always ordered from least to greatest. It is written as below.

If we look at the y-values on the graph, we get -2,-3,0,and 3. If we order them, we get -3,-2,0, and 3.

This means our answer is B.

I hope this helps!

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Please help if possible its important!!

Otrada [13]

Answer:

600 ft³

Step-by-step explanation:

Base area: B

Base is triangle

base = 12 ft and height = 5 ft

Area of triangle = B = $\frac{1}{2}* \ base * \ height$

$= \frac{1}{2}*12 * 5\\\\= 6 * 5\\\\= 30 \ ft^{2}$

H = 20 ft

Volume = B * H

= 30 * 20

= 600 ft³

7 0

3 years ago

The graph of a quadratic function has roots at -3 and -7 and a vertex at (-5, 4). What is the equation of the function in vertex

melomori [17]

Answer:

y = -(x + 5)² + 4

Step-by-step explanation:

The roots are -3 and -7, so:

y = a (x + 3) (x + 7)

Distribute and complete the square:

y = a (x² + 10x + 21)

y = a (x² + 10x + 25 − 4)

y = a (x² + 10x + 25) − 4a

y = a (x + 5)² − 4a

The vertex is (-5, 4), so a = -1.

y = -(x + 5)² + 4

3 0

3 years ago

What is the answer to -7(-3n 2)?

sergey [27]

21n+14 you have to find the value of n

8 0

3 years ago

John's car gets 25.3 miles per gallon. How many<br> miles can he drive on 11.7 gallons of gas?

bekas [8.4K]

Answer:

Multiply 25.3 by 11.7, to get the answer of 296.01

8 0

3 years ago

F=e−yi−xe−yj is conservative. find a scalar potential f and evaluate the line integral over any smooth path c connecting a(0,0)

Alexxx [7]

If $\mathbf F$ is conservative, then there is a scalar function $f$ such that

$\nabla f(x,y)=\mathbf F(x,y)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j=e^{-y}\,\mathbf i-xe^{-y}\,\mathbf j$

Setting the first components equal to one another, we can integrate both sides to find

$\dfrac{\partial f}{\partial x}=e^{-y}\implies f(x,y)=xe^{-y}+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial f}{\partial y}=-xe^{-y}+\dfrac{\mathrm dg}{\mathrm dy}=-xe^{-y}$
$\implies\dfrac{\mathrm dg}{\mathrm dy}=0$
$\implies g(y)=C$

so that

$f(x,y)=xe^{-y}+C$

By the fundamental theorem of calculus, we have that

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(1,1)-f(0,0)=\frac1e$

4 0

3 years ago