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Elden [556K]
3 years ago
5

Please help!

Mathematics
1 answer:
Anna [14]3 years ago
3 0

Answer:

-1 and -2

Step-by-step explanation:

Point of intersection of the two graphs are the solution to the equation.

The point of intersections : -1 , -2

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The whale swims down 300 feet to feed. What is the whale elevation now
Luba_88 [7]

Answer:

If the whale swims 300 feet below the surface it would be -300 feet below the sea level

Step-by-step explanation:

5 0
3 years ago
The points (6, 9) and (0, 6) fall on a particular line. What is its equation in slope-intercept form?
SashulF [63]

Answer:

y = \frac{1}{2}x + 6

Step-by-step explanation:

Let \ (x _1  y _ 1 ) = ( 6 , 9 )  \ and  \ (x _ 2 , y _ 2 ) \  = ( 0 , 6 )\\\\

Step 1 : Find slope, m

              slope, m = \frac{y _ 2 - y _ 1}{x_ 2 - x_ 1 } = \frac{6 - 9}{0 - 6 } = \frac{-3}{-6} = \frac{1}{2}

Step 2 : Find the equation

                (y - y _ 1 ) = m ( x - x_ 1) \\\\( y - 9 ) = \frac{1}{2} (x - 6 ) \\\\y = \frac{1}{2}x - 3 + 9 \\\\y = \frac{1}{2}x + 6

3 0
3 years ago
Read 2 more answers
Find the measure of angle A when two angles are congruent?
erica [24]

m∡A = 70º

1) Considering that the Sum of the interior angles within a triangle is always 180º

2) We can write the following, and solve for x

x+80 + x +80 +40 = 180

2x +160 +40 = 180

2x + 200 = 180

2x =180-200

2x= -20

x=-10

3) So since angle A = x +80, we can plug it into that the value of x

m∡A = x +80º

m∡A=-10+80

m∡A = 70º

4 0
1 year ago
Which statement is true?<br> GRAPHA<br> GRAPH B
WARRIOR [948]

Answer:

Graph B is correct

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Sphere A is similar to Sphere B. The scale factor of the lengths of the radii of Sphere A to Sphere B is 1 to 4. Sphere A has th
AleksAgata [21]

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{\textit{sphere A}}{\textit{sphere B}}\qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}\qquad \qquad \stackrel{\stackrel{sides'}{ratio}}{\cfrac{1}{4}}=\stackrel{\stackrel{volumes'}{ratio}}{\cfrac{\sqrt[3]{288}}{\sqrt[3]{v}}}\implies \cfrac{1}{4}=\sqrt[3]{\cfrac{288}{v}}\implies \left( \cfrac{1}{4} \right)^3=\cfrac{288}{v} \\\\\\ \cfrac{1^3}{4^3}=\cfrac{288}{v}\implies \cfrac{1}{64}=\cfrac{288}{v}\implies v=18432

3 0
4 years ago
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