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3 years ago

What is the midpoints of a segment whose endpoints are (-3, 8) and (7, -2) ?

1 answer:

3 0

$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$M = (\frac{7 - (-3)}{2}, \frac{-2 + 8}{2})$
$M = (\frac{7 - 3}{2}, \frac{6}{2})$
$M = (\frac{4}{2}, 3)$
$M = (2, 3)$

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If the point (x,square root 3/2) is on the unit circle, what. is x?

galina1969 [7]

Answer:

$x=\pm\frac{1}{2}$

Step-by-step explanation:

On a unit circle, $(x,y)=(\cos\theta,\sin\theta)$ , so $\sin\theta=\frac{\sqrt{3}}{2}$ in this case. If you look at the attached circle, the only time that the y-coordinate is $\frac{\sqrt{3}}{2}$ is when $x=-\frac{1}{2}$ and $x=\frac{1}{2}$ , which correspond to angles of $\theta=\frac{2\pi}{3}$ and $\theta=\frac{\pi}{3}$ respectively

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2 years ago

Triangle Proofs Please help

oksian1 [2.3K]

Explanation:

There may be a more direct way to do this, but here's one way. We make no claim that the statements used here are on your menu of statements.

<u>Statement</u> . . . . <u>Reason</u>

2. ∆ADB, ∆ACB are isosceles . . . . definition of isosceles triangle

3. AD ≅ BD

and ∠CAE ≅ ∠CBE . . . . definition of isosceles triangle

4. ∠CAE = ∠CAD +∠DAE

and ∠CBE = ∠CBD +∠DBE . . . . angle addition postulate

5. ∠CAD +∠DAE ≅ ∠CBD +∠DBE . . . . substitution property of equality

6. ∠CAD +∠DAE ≅ ∠CBD +∠DAE . . . . substitution property of equality

7. ∠CAD ≅ ∠CBD . . . . subtraction property of equality

8. ∆CAD ≅ ∆CBD . . . . SAS congruence postulate

9. ∠ACD ≅ ∠BCD . . . . CPCTC

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4 0

2 years ago

Which of the following graphs shows a pair of lines that represents the equations with the solution (3, −6)?

Deffense [45]

Answer:

the last one

Step-by-step explanation:

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3 years ago

Evaluate the double integral. 2y2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1) D

zubka84 [21]

Answer:

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

Step-by-step explanation:

From the image attached below;

We need to calculate the limits of x and y to find the double integral

We will notice that y varies from 1 to 2

The line equation for (0,1),(1,2) is:

$y-1 = \dfrac{2-1}{1-0}(x-0)$

y - 1 = x

The line equtaion for (1,2),(4,1) is:

$y-2 = \dfrac{1-2}{4-1}(x-1) \\ \\ y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = (x -1)

-3y + 6 = x - 1

-x = 3y - 6 - 1

-x = 3y - 7

x = -3y + 7

This implies that x varies from y - 1 to -3y + 7

Now, the region D = {(x,y) | 1 ≤ y ≤ 2, y - 1 ≤ x ≤ -3y + 7}

The double integral can now be calculated as:

$\iint _D y^2 dA= \int ^2_1 \int ^{-3y +7}_{y-1} \ 2y ^2 \ dx \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[ 2xy ^2 \bigg]^{-3y+7}_{y-1} \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[2(-3y+7)y^2-2(y-1)y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-6y^3 +14y^2 -2y^3 +2y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-8y^3 +16y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \bigg[-8(\dfrac{y^4}{4}) +16(\dfrac{y^3}{3})\bigg ] ^2_1$

$\iint _D y^2 dA= \bigg[-8(\dfrac{16}{4}-\dfrac{1}{4}) +16(\dfrac{8}{3}-\dfrac{1}{3})\bigg ]$

$\iint _D y^2 dA= \bigg[-8(\dfrac{15}{4}) +16(\dfrac{7}{3})\bigg ]$

$\iint _D y^2 dA= -30 + \dfrac{112}{3}$

$\iint _D y^2 dA= \dfrac{-90+112}{3}$

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

4 0

3 years ago

CAN SOMEONE HELP?????????????????????????????????

Zanzabum

Answer:

5/7 and -7/5

Step-by-step explanation:

slope is - 7/5

if line is written as Ax + By + C = 0

the slope is -A/B

if you work that equation you have

By = - Ax - C

y = (-A/B) × x - C/B

and term multiplying the x is the slope in case you had doubts about the - A/B

the multiplication of the slopes of 2 lines that are perpendicular is - 1

x × -7/5 = -1

x = 5/7

the slope of 2 lines that are parallels is the same, so slope = -7/5

5 0

3 years ago

Read 2 more answers