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8090 [49]
3 years ago
11

Which expression is equivalent to -51 - (-60)−51−(−60)minus, 51, minus, left parenthesis, minus, 60, right parenthesis?

Mathematics
1 answer:
djyliett [7]3 years ago
5 0

Answer:

it's B.

Step-by-step explanation:

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Find the value of x
Svet_ta [14]

Answer:

150^o

Step-by-step explanation:

<u>Step 1:  Distribute the plus</u>

<u />(x + 15)^o + (135 - x)^o

x + 15 + 135 - x

<u>Step 2:  Combine like terms</u>

<u />x + 15 + 135 - x

(x - x)^o + (15 + 135)^o

0 + 150^o

150^o

Answer:  150^o

7 0
3 years ago
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If KL = LM and KJ = JM, find m ∠KJL<br><br> please, i really need your help!
umka2103 [35]
The triangles are congruent by SAS, so the two angles with 7x and 3x+16 are equal. So

7x=3x+16
4x=16
x=4

So KJL=7(4)=28.
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3 years ago
This isn't a math question. Hellooo!!! Can someone pls help me??? how does racial discrimination relate to economic statificatio
borishaifa [10]
Im confused on what are you asking
6 0
3 years ago
Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter
Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and  \sqrt{8^{2} - x^{2} }or  \sqrt{64 - x^{2} }

thus tan(arcsin(x/8))  = \frac{x}{\sqrt{64 - x^{2} }}  =

Therefore  ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is \sqrt{64 - x^{2} } and hypothenuse = 8 we have \sqrt{64 - x^{2} }/8

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

4 0
3 years ago
What is the following product? sqrt 12*sqrt 18
uysha [10]
------------------
√12*√18

2√3√18

2√3 x 3√2

2 x 3√3 x 2

2 x 3√6

6√6 or 14.696938
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6 0
3 years ago
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