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3 years ago

Mhanifa please help! I will mark brainliest! I really don’t know how to do this! Random answers will be reported

Mathematics

2 answers:

Butoxors [25]3 years ago

8 0

Answer:

<u>Supplementary angles sum to 180°</u>

9y + 46 + 7y - 10 = 180
16y + 36 = 180
16y = 144
y = 144/16
y = 9

<u>Complementary angles sum to 90°</u>

Complement of ∠M = 90° - 32.6° = 57.4°

<u>Angle addition</u>

m∠TOP = m∠TOM + m∠MOP
m∠TOP = 12.5° + 16.5°
m∠TOP = 29°

BartSMP [9]3 years ago

6 0

I believe the value of y is -28

also believe m
25 is 57.4

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vlabodo [156]

Answer:

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3 years ago

Drako found an emerald in a cave at a depth between Negative one-half and Negative 1 and two-thirds meters. Which number could r

8090 [49]

Options :

Negative 2 and one-third meters

Negative three-fourths meters

Negative one-fourth meters

Negative 1 and

StartFraction 5 Over 6 EndFraction meters

Answer: Negative three-fourths meters

Step-by-step explanation:

Given the following :

Depth Location of Emerald :

Between - 1/2 and - 1 2/3

The answer will be the option which lies between the two depth points.

-1/2 = - 0.5 ( upper bound)

-1 2/3 = - 1.667 ( lower bound)

The option which lies inbetween :

- 1.667 and - 0.5

Taking each option :

Negative 2 and one-third meters = - 2 1/3 = - 2.33

Negative three-fourths meters = - 3/4 = - 0.75

Negative one-fourth meters = - 1/4 = - 0.25

Negative 1 and StartFraction 5 Over 6 EndFraction meters = - 1 5/6 = - 1.833

Hence, the only depth position which lies inbetween (-1.667 and - 0.5) is - 0.75

4 0

3 years ago

Read 2 more answers

Which solution to the equation 1/x-1=x-2/2x^2-2 is extraneous?

olga2289 [7]

<span>An extraneous solution is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.

Given the equation:
$\frac{1}{x-1} = \frac{x-2}{2x^2-2} \\ \\ 2x^2-2=(x-1)(x-2)=x^2-3x+2 \\ \\ x^2+3x-4=0 \\ \\ (x-1)(x+4)=0 \\ \\ x-1=0 \ or \ x+4=0 \\ \\ x=1 \ or \ x=-4$

From the equation, it is obtained that x = 1 and x = -4 are the solutions of the equation.

Now, we substitute x = 1 into the equation as follows:
</span><span> $\frac{1}{1-1} = \frac{1-2}{2(1)^2-2} \\ \\ \frac{1}{0} = \frac{-1}{2-2} \\ \\ \frac{1}{0} = \frac{-1}{0}$

As can be see, for x = 1, the equation is undefined.

Now, we substitute x = -4 into the equation as follows:
$\frac{1}{-4-1} = \frac{-4-2}{2(-4)^2-2} \\ \\ \frac{1}{-5} = \frac{-6}{2(16)-2} = \frac{-6}{32-2} = \frac{-6}{30} \\ \\ - \frac{1}{5} =- \frac{1}{5}$

It can be seen that x = -4 is a valid solution of the orignal equation.

Therefore, x = 1 is an extraneous solution to the equation
$\frac{1}{x-1} = \frac{x-2}{2x^2-2}$ </span>

8 0

3 years ago

(4x3 + x2 - 9) - (- 5x3 + 7x - 16)

scoray [572]

Answer:

34-5x

Step-by-step explanation:

(12+2x-9)-(-15+7x-16)

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3+2x+31-7x

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3 years ago

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If (82P = 84, what is the value of p?

nata0808 [166]

Answer:

p=42/41 or 1 an 1/41

Step-by-step explanation:

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3 years ago