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1 year ago

PLEASE HELPPPPPPPPPPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Serga [27]1 year ago

7 0

The value of the variable g is 6.9

<h3>Square root</h3>

From the question, we are to determine the value of the unknown variable in the given equation

The given equation is

g² = 47

The unknown variable is g

Solving for g

g² = 47

Note that the variable g is squared

To solve for the variable g in the given equation, we will take the square roots of both sides of the equation

So that,

g = √47

g = 6.8556546

g ≈ 6.9 (to the nearest tenth)

Hence, the value of the variable g is 6.9

Learn more on Square root here: brainly.com/question/12186958

#SPJ1

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Each of the p children at a park receives at least 3 balloons. If the children receive 150 balloons in all, find the maximum num

STALIN [3.7K]

The answer i think is 50

3 0

3 years ago

Read 2 more answers

What is the measure of B (2n + 15) and D (3n - 5)? If ABCD is a parrallelogram

katovenus [111]

Answer : The value of angle B and angle D is 25⁰ and 35⁰ respectively.

Step-by-step explanation :

As we know that the opposite angles are equal in parallelogram.

According to the given figure,

∠A = ∠C

and

∠B = ∠D

Given:

∠B = (3n - 5)⁰

∠D = (2n + 15)⁰

From this we conclude that:

∠B = ∠D

(3n - 5)⁰ = (2n + 15)⁰

3n - 5⁰ = 2n + 15⁰

3n - 2n = 15⁰ - 5⁰

1n = 10⁰

n = 10⁰

∠B = (3n - 5)⁰ = (3×10 - 5)⁰ = 25⁰

∠D = (2n + 15)⁰ = (2×10 + 15)⁰ = 35⁰

Therefore, the value of angle B and angle D is 25⁰ and 35⁰ respectively.

5 0

3 years ago

Please help me<br><br> Find the value of x

kvv77 [185]

Answer:

x = 5 and y = 22

Step-by-step explanation:

See attached image.

6 0

2 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago

12.09 rounded to the first decimal place

bulgar [2K]

12.0 is the answer I think

4 0

3 years ago