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2 years ago

The following table gives the statements and justifications for solving the equation 4(x - 7) + 14 =

Mathematics

1 answer:

evablogger [386]2 years ago

4 0

Answer:

From top to bottom:

Simplify

Combine like terms

Subtract 3x from both sides

Add 14 on both sides

Step-by-step explanation:

Hope this helps

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Find the Area. Simplify your answer.

Vaselesa [24]

Answer:

Area = $9x^2$ square units

Step-by-step explanation:

Key skills needed: Area of a parallelogram

1) This shape is a parallelogram, and the way to find the area of a parallelogram is:

$A = bh$

A = area

b = base (the side at the bottom

h = the height ( the length that runs from one side to another and makes a right angle)

2) The base (b) would be 3x. The height (h) would also be 3x

3) This means A = $3x(3x)$

3 times 3 is 9 --> and x times x is $x^2$

Therefore the area is --> $9x^2$

Hope you understood and have a nice day!! :D

5 0

3 years ago

Someone tell me what (GJ) is l please!!

skad [1K]

Answer:

that box means the angle is 90 degrees.

Step-by-step explanation:

if you need anymore help just message me

4 0

3 years ago

A grey whale’s heart beats 24 times in 3 min. How many times does it beat in a day?

zepelin [54]

In i hour it would be
60/3=20
20x24=480
480x24=11520.
11520 is your answer! :)
Hope this helps!!

~Summer

7 0

3 years ago

Read 2 more answers

What is the equation to the line that goes through the point

DochEvi [55]

Step-by-step explanation:

perpendicular slope is negative reciprocal. -1/-1 = 1

y - 4 = x +3

y = x + 7

5 0

3 years ago

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in fourth quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in 4th quadrant so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

3 years ago