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2 years ago

Please help and thank you!

Mathematics

1 answer:

lukranit [14]2 years ago

6 0

9514 1404 393

Answer:

base area: 21 ft²
volume 105 ft³

Step-by-step explanation:

The area of a triangle is given by the formula ...

A = (1/2)bh

The base and height of a right triangle can be taken to be the leg lengths, since each is perpendicular to the other.

A = (1/2)(6 ft)(7 ft) = 21 ft² . . . . base area

The volume is the product of the base area and the height:

V = Bh

V = (21 ft²)(5 ft) = 105 ft³ . . . . volume

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What is the value of m? 0.61m−1.51m=9 Enter your answer in the box. m =

sergij07 [2.7K]

Answer:

m = -10

Step-by-step explanation:

0.61m - 1.51m = 9

-0.90m = 9

mutiply both sides by -1.111111111111, twelve 1's after the decimal point

also to find what number u need to multiply to make -0.9m to m, divide 0.9 by 1 and that equals 1.111111111111 so yea a lot of ones haha

m = -10

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2 years ago

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PLEASE HELP!!!!

Lunna [17]

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

if "a" is positive, then opens UP
if "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

the Axis Of Symmetry (x-value) is: x = $\frac{-b}{2a}$
y-value is found by plugging in the AOS for "x" in the equation

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1) y = x² + 11x + 24

a = +1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$

y = $(-\frac{11}{2})^{2}$ + 11 $(-\frac{11}{2} )$ + 24 = $-\frac{25}{4}$

vertex $(-\frac{11}{2}$ , $-\frac{25}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

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2) y = -x² - 6x - 8

a = -1 so the parabola opens DOWN

x = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3

y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1

vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

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3) y = x² - 2x + 3

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1

y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). There are no x-intercepts

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4) y = x² + 4x + 4

a = 1 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2

y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0

vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). The x-intercept is the vertex.

Compared to the other four graphs, this is most likely the equation for the rain gauge!

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a = +3 so the parabola opens UP

x = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$

y = 3 $(-\frac{7}{2})^{2}$ + 21 $(-\frac{7}{2} )$ + 30 = $-\frac{27}{4}$

vertex $(-\frac{7}{2}$ , $-\frac{27}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

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3 years ago