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crimeas [40]
3 years ago
13

Please help, solve for x?

Mathematics
2 answers:
STatiana [176]3 years ago
8 0

Answer:

x = 3

Step-by-step explanation:

Assuming QTSR is a parallelogram, then segment QT is of equal length to segment SR. Solve for x by writing the equation 9x - 3 = 24

Eva8 [605]3 years ago
3 0

Answer:

x = 3

Step-by-step explanation:

24 = 9x -3

(add 3)

27 = 9x

(÷9)

x = 3

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Help please i need this done by tmr !
Nataly [62]

Answer:

see explanation

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

sum the 3 angles and equate to 180

5x + 7 + 4x + 2 + 90 = 180 , that is

9x + 99 = 180 ( subtract 99 from both sides )

9x = 81 ( divide both sides by 9 )

x = 9

Then

∠ ACB = 4x + 2 = 4(9) + 2 = 36 + 2 = 38°

∠ BAC = 5x + 7 = 5(9) + 7 = 45 + 7 = 52°

-----------------------------------------------------------------------

Since the triangles are congruent then corresponding angles are congruent , so

∠ B = ∠ E

6x + 10 = 70 ( subtract 10 from both sides )

6x = 60 ( divide both sides by 6 )

x = 10

5 0
3 years ago
If someone could answer this correctly I will Venmo you $15
Gelneren [198K]

Answer:

<u><em>0-20</em></u>

Step-by-step explanation:

6 0
3 years ago
What is the length of segment AE?
nexus9112 [7]

Answer:

AE=\frac{50}{3}\ units

Step-by-step explanation:

step 1

In the right triangle ABC

Applying the Pythagoras Theorem fin the hypotenuse AC

AC^{2} =AB^{2}+BC^{2}

substitute

AC^{2} =6^{2}+8^{2}

AC^{2} =100

AC =10\ units

step 2

we know that

If two figures are similar, then the ratio of its corresponding sides is equal

so

\frac{AB}{CD}=\frac{AC}{CE}

substitute and solve for CE

\frac{6}{4}=\frac{10}{CE}\\ \\CE=4*10/6\\ \\CE=\frac{20}{3}\ units

step 3

Find the length of segment AE

AE=AC+CE

substitute the values

AE=10\ units+\frac{20}{3}\ units=\frac{50}{3}\ units

6 0
3 years ago
Find the volume of the square based pyramid.
frozen [14]
A = 1/3 * s^2 * h where s is a side of the base and h is the height.
4 0
3 years ago
Read 2 more answers
3. Which of the following quadratic equations has no solution? A) 0 = −2(x − 5)2 + 3 B) 0 = −2(x − 5)(x + 3) C) 0 = 2(x − 5)2 +
klemol [59]

Answer:

D) 0 = 2(x + 5)(x + 3)

Step-by-step explanation:

Which of the following quadratic equations has no solution?

We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.

A) 0 = −2(x − 5)2 + 3

0 = -2(x - 5) × 5

0 = (-2x + 10) × 5

0 = -10x + 50

10x = 50

x = 50/10

x = 5

Option A has a solution of 5

B) 0 = −2(x − 5)(x + 3)

Take each of the factors and equate them to zero

-2 = 0

= 0

x - 5 = 0

x = 5

x + 3 = 0

x = -3

Option B has a solution by one of its factors as a positive value of 5

C) 0 = 2(x − 5)2 + 3

0 = 2(x - 5) × 5

0 = (2x -10) × 5

0 = 10x -50

-10x = -50

x = -50/-10

x = 5

Option C has a solution of 5

D) 0 = 2(x + 5)(x + 3)

Take each of the factors and equate to zero

0 = 2

= 0

x + 5 = 0

x = -5

x + 3 = 0

x = -3

For option D, all the values of x are 0, or negative values of -5 and -3.

Therefore the Quadratic Equation for option D has no solution.

3 0
3 years ago
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