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11Alexandr11 [23.1K]
3 years ago
12

-6 + (-2) + 5= ? Help would really be appreciated!! :)

Mathematics
2 answers:
lozanna [386]3 years ago
6 0

Answer:

-6 + (-2) + 5 = -3

(Hope this helps! Btw, I am the first to answer. Brainliest pls! :D)

kvasek [131]3 years ago
4 0

Answer:

-3

Step-by-step explanation:

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Find the value of x.
son4ous [18]

Answer:

18.0

Step-by-step explanation:

==>Given:

Triangle with sides, 16, 30, and x, and a measure of an angle corresponding to x = 30°

==>Required:

Value of x to the nearest tenth

==>Solution:

Using the Cosine rule: c² = a² + b² - 2abcos(C)

Let c = x,

a = 16

b = 30

C = 30°

Thus,

c² = 16² + 30² - 2*16*30*cos 30°

c² = 256 + 900 - 960 * 0.8660

c² = 1,156 - 831.36

c² = 324.64

c = √324.64

c = 18.017769

x ≈ 18.0 (rounded to nearest tenth)

8 0
3 years ago
Would anybody give the right answer
BARSIC [14]

Answer:

i think so

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Which of the following best describes the equation below?<br><br> y = |x| + 5
sleet_krkn [62]

Answer:

On a graph, you go up 5.

4 0
3 years ago
Given the following table of values for f(x), find f(2).
AleksandrR [38]

Step-by-step explanation:

Given F(x) is 5 10 8 5 7 14 where x is - 1 0 1 2 3 9,

the value of F(2) is 202,478

3 0
2 years ago
How to find an exponential function with two points?
BabaBlast [244]
Well u have a basic equation y=a • b^x
a is the initial amount
b is the growth factor
and x is the exponent

the first step is to make a chart with ur two points for example
(0,4) and (2,16)
so ur chart will be
x. | y.
0. | 4
2. | 16
next you find the difference between the x side so 0 to 2 is +2
then find the difference between the y side so 4 to 16 is +12
then put it into a fraction with y over x or y/x so 12/2 then simplified 6/1 or just 6.

6 is the growth factor
and to find a u have to go on the y column and find the first number so
a is 4
x is still x because it's 0 but if it was 2 then it would be x-2 so it can cancel
so the answer would be y=4 • 6^x
6 0
3 years ago
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