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Neko [114]
3 years ago
10

Help please :/ jejsjsj

Mathematics
2 answers:
Alla [95]3 years ago
6 0

Answer:

-6

-21

12

Step-by-step explanation:

can i get brainly

Colt1911 [192]3 years ago
3 0

Answer:

- 6, - 21, 12

Step-by-step explanation:

(1)

2x - 7 = - 19 ( add 7 to both sides )

2x = - 12 ( divide both sides by 2 )

x = - 6

(2)

\frac{x}{-3} + 5 = 12 ( subtract 5 from both sides )

\frac{x}{-3} = 7 ( multiply both sides by - 3 to clear the fraction )

x = - 21

(3)

\frac{3}{4} x - 5 = 4 ( add 5 to both sides )

\frac{3}{4} x = 9 ( multiply both sides by 4 to clear the fraction )

3x = 36 ( divide both sides by 3 )

x = 12

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Help! what is 302938*4043
bonufazy [111]

Answer:

1,224,778,334

Step-by-step explanation:

302938 \times 4043

= 1224778334

Hope it helps you!!

#IndianMurga(. ❛ ᴗ ❛.)

6 0
3 years ago
Read 2 more answers
Consider the points A(5, 3t+2, 2), B(1, 3t, 2), and C(1, 4t, 3). Find the angle ∠ABC given that the dot product of the vectors B
Vilka [71]

Answer:

66.42°

Step-by-step explanation:

<u>Given:</u>

A(5, 3t+2, 2)

B(1, 3t, 2)

C(1, 4t, 3)

BA • BC = 4

Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

BC = OC - OB = (1-1, 4t-3t, 3-2) = (0, t, 1)

Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

2 = t

t = 2

Now we know that:

BA = (4, 2, 0)

BC = (0, 2, 1)

Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

BA • BC = 4

|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}

|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}

2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4

10\cos{(m\angle{ABC})} = 4

\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}

m\angle{ABC} = cos^{-1}{(\frac{2}{5})}

m\angle{ABC} = 66.4218^{\circ}

Rounded to two decimal places:

m\angle{ABC} = 66.42^\circ

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2 years ago
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3 years ago
How to solve the 4a.plz
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