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3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP ME!!!!!!!

Mathematics

1 answer:

professor190 [17]3 years ago

5 0

1: x=-2 2: x=-3 i did send a pic hopefully you can see it is not I will type in comments how to solve it

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What decimal is equivalent to the fraction and percent below?

REY [17]

.10 if it’s out of 1

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3 years ago

Read 2 more answers

2. Point A is located at (-8, -3) on the coordinate grid. If it is

musickatia [10]

Answer:

(-6, 6).

Step-by-step explanation:

The x-coordinate increases by 2 units and the y-coordinate by 9 units:

That is ( - 8 + 2, -3 + 9)

= (-6, 6).

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3 years ago

Find the measure of each angle.

Galina-37 [17]

<u>Given</u>:

Given that the isosceles trapezoid JKLM.

The measure of ∠K is 118°

We need to determine the measure of each angle.

<u>Measure of ∠L:</u>

By the property of isosceles trapezoid, we have;

$\angle K+\angle L=180^{\circ}$

$118^{\circ}+\angle L=180^{\circ}$

$\angle L=62^{\circ}$

Thus, the measure of ∠L is 62°

<u>Measure of ∠M:</u>

By the property of isosceles trapezoid, we have;

$\angle L \cong \angle M$

Substituting the value, we get;

$62^{\circ}=\angle M$

Thus, the measure of ∠M is 62°

<u>Measure of ∠J:</u>

By the property of isosceles trapezoid, we have;

$\angle J \cong \angle K$

Substituting the value, we get;

$\angle J =118^{\circ}$

Thus, the measure of ∠J is 118°

Hence, the measures of each angles of the isosceles trapezoid are ∠K = 118°, ∠L = 62°, ∠M = 62° and ∠J = 118°

4 0

3 years ago

Graph f(x)= 1/x-2 include asymptotes, at least 5 points, and neatly sketch the branches

riadik2000 [5.3K]

Answer:

I attached the answer below.

Step-by-step explanation:

I recommend using Desmos, that's what I used. It helps graph and plot points.

Hope this helps!!

7 0

2 years ago

|. Identify the following Pōints of each values.Write your ans

Dmitry_Shevchenko [17]

<h2>✒️VALUE</h2>

$\\ \quad \begin{array}{c} \qquad \bold{Distance \: \green{ Formula:}}\qquad\\ \\ \boldsymbol{ \tt d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} \end{array}\\ \begin{array}{l} \\ 1.)\: \bold{Given:}\: \begin{cases}\tt D(- 5,6), E(2.-1),\textsf{ and }F(x,0) \\ \tt DF = EF \end{cases} \\ \\ \qquad\bold{Required:}\:\textsf{ value of }x \\ \\ \qquad \textsf{Solving for }x, \\ \\ \tt \qquad DF = EF \\ \\ \implies\small \tt{\sqrt{(x -(- 5))^2 + (0 - 6)^2} = \sqrt{(x - 2)^2 + (0 - (-1))^2}} \\ \\ \implies\tt\sqrt{(x + 5)^2 + 36 } = \sqrt{(x - 2)^2 + 1 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt (x + 5)^2 + 36 = (x - 2)^2 + 1 \\ \\ \implies\tt x^2 + 10x + 25 + 36 = x^2 - 4x + 4 + 1 \\ \\ \implies \tt x^2 + 10x + 61 = x^2 - 4x + 5 \\ \\ \implies\tt10x + 4x = 5 - 61 \\ \\ \implies\tt14x = -56 \\ \\ \implies \red{\boxed{\tt x = -4}}\end{array} \\ \\ \\ \\\begin{array}{l} \\ 2.)\: \bold{Given:}\: \begin{cases}\tt P(6,-1), Q(-4,-3),\textsf{ and }R(0,y) \\ \tt PR = QR \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }y \\ \\ \qquad\textsf{Solving for }y, \\ \\ \qquad\tt PR = QR \\ \\ \implies \tt\small{\sqrt{(0 - 6)^2 + (y - (-1))^2} = \sqrt{(0 - (-4))^2 + (y - (-3))^2}} \\ \\ \implies\tt\sqrt{36 + (y + 1)^2} = \sqrt{16 + (y + 3)^2 } \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies \tt \: 36 + (y + 1)^2 = 16 + (y + 3)^2 \\ \\ \implies\tt 36 + y^2 + 2y + 1 = 16 + y^2 + 6y + 9 \\ \\ \implies \tt \: y^2 + 2y + 37 = y^2 + 6y + 25 \\ \\ \implies \tt \: 2y - 6y = 25 - 37 \\ \\ \implies \tt -4y = -12 \\ \\ \implies\red{\boxed{ \tt y = 3}} \end{array} \\ \\ \\ \begin{array}{l} \\ 3.)\: \bold{Given:}\: \begin{cases}\: A(4,5), B(-3,2),\textsf{ and }C(x,0) \\ \: AC = BC \end{cases} \\ \\ \bold{Required:}\:\textsf{ value of }x \\ \\ \qquad\textsf{Solving for }x, \\ \\ \qquad\tt AC = BC \\ \\ \implies\tt\small{\sqrt{(x - 4)^2 + (0 - 5)^2} = \sqrt{(x - (-3))^2 + (0 - 2)^2}} \\ \\ \implies\tt\sqrt{(x - 4)^2 + 25} = \sqrt{(x + 3)^2 + 4} \\ \\ \textsf{Squaring both sides yields} \\ \\ \implies\tt\:(x - 4)^2 + 25 = (x + 3)^2 + 4 \\ \\ \implies\tt\:x^2 - 8x + 16 + 25 = x^2 + 6x + 9 + 4 \\ \\ \implies\tt\:x^2 - 8x + 41 = x^2 + 6x + 13 \\ \\ \implies\tt-8x - 6x = 13 - 41 \\ \\\implies\tt -14x = -28 \\ \\ \implies\red{\boxed{\tt\:x = 2}} \end{array}$

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2 years ago