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3 years ago

10

What is the value of x? sin(x+22)°=cos(2x−7)°

1 answer:

Snezhnost [94]3 years ago

5 0

Remember that $\sin(\theta) = \cos(90^\circ - \theta)$

Thus, we can solve for x.
x + 22 = 90 - (2x - 7)
x + 22 = 97 - 2x
3x + 22 = 97
3x = 75
x = 25

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(4/2)4 + (1 + 3)?<br><br> I really need help!

Katyanochek1 [597]

$\huge\text{Hey there!}$

$\large\text{Just do PEMDAS}$

$\large\text{Parentheses}$

$\large\text{Exponents}$

$\large\text{Multiplication}$

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$\large\text{Addition}$

$\large\text{Subtraction}$

$\mathsf{(\dfrac{4}{2})(4) + (1+3)}$

$\mathsf{\dfrac{4}{2} = 2}$

$\mathsf{2(4) + (1 + 3)}$

$\mathsf{2(4) = 8}$

$\mathsf{8 + (1 + 3)}$

$\mathsf{1 + 3 = 4}$

$\mathsf{8 + 4 = 12}$

$\boxed{\boxed{\large\text{Answer: \bf \huge12 }}}\huge\checkmark$

$\text{Good luck on your assignment and enjoy your day! }$

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6 0

3 years ago

The diffrence between 2 numbers is 1 and their product Is 132

il63 [147K]

Answer:

x = 12

y = 11

For any clarification, Please contact me in the comment section :)

I Hope It Helps .

Step-by-step explanation:

$Let -the-unkonown- numbers-be-x-and-y \\x-y = 1 (1)\\xy = 132 (2)\\Make-x-subject-of-formula-in-equation1 \\x = 1+y\\Substitute , 1+y for- x -in -equation ,2\\y(1+y) = 132\\y + y^{2} =132\\y^{2} +y-132=0\\Solve- the-quadratic-equation.\\y = 11\\substitute- 11 -for- y- in- equation -1- 0r- 2\\x -y = 1\\x-11 = 1\\x = 1+11\\x = 12$

6 0

3 years ago

At camp, ¼ of the children are divided into 3 equal groups for crafts. What fraction of the children at camp are in EACH group?

zloy xaker [14]

Answer:

3/4?

Step-by-step explanation:

maybe not but i think this is it

7 0

3 years ago

Read 2 more answers

You have 5,814 feet of rope that you need to cut into 27-foot sections. How many sections of rope will you have?

natta225 [31]

5814/27 = 215.33

215 x 27 = 5805

5814-5805 = 9

Answer is B 215 sections with 9 feet left over

8 0

2 years ago

Given the graph of a degree 3 polynomial below, complete the table of values for either the x-value of a zero, or the multiplici

leva [86]

The roots or zeros of a polynomial are the points it crosses the x-axis.

The roots of the polynomial in order from least to greatest are -5, 0 and 5

From the graph (see attachment), we have the following highlights

<em>The graph cross the x-axis at x = -5</em>
<em>The graph cross the x-axis at x = 0</em>
<em>The graph cross the x-axis at x = 5</em>

The points at which the graph crosses the x-axis are the roots of the graph.

So, the roots are: -5, 5 and 5

Read more about the roots of a polynomial at:

brainly.com/question/7921963

8 0

2 years ago