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4 years ago

Are triangles PRU and STQ congruent?

Mathematics

1 answer:

san4es73 [151]4 years ago

7 0

No, triangles PRU and STQ are not congruent. Even though their angles are the same, different triangles can be constructed out of triangles with the same angles. Looking at the picture, one can also see the difference in size between the two triangles.

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Are these triangles congruent? If so, name the triangle congruence method.

mars1129 [50]

Answer:

The answer you chose is correct

Step-by-step explanation:

4 0

3 years ago

Solve each inequality, and then drag the correct solution graph to the inequality.

Nesterboy [21]

The correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

(NOTE: The graphs are labelled A, B and C from left to right)

For the first inequality,

$4(9x-18)>3(8x+12)$

First, clear the brackets,

$36x-72>24x+36$

Then, collect like terms

$36x-24x>36+72\\12x >108$

Now divide both sides by 12

$\frac{12x}{12} > \frac{108}{12}$

∴ $x > 9$

For the second inequality

$-\frac{1}{3}(12x+6) \geq -2x +14$

First, clear the fraction by multiplying both sides by 3

$3 \times[-\frac{1}{3}(12x+6)] \geq3 \times( -2x +14)$

$-1(12x+6) \geq -6x +42$

Now, open the bracket

$-12x-6 \geq -6x +42$

Collect like terms

$-6 -42\geq -6x +12x$

$-48\geq 6x$

Divide both sides by 6

$\frac{-48}{6} \geq \frac{6x}{6}$

$-8\geq x$

∴ $x\leq -8$

For the third inequality,

$1.6(x+8)\geq 38.4$

First, clear the brackets

$1.6x + 12.8\geq 38.4$

Collect likes terms

$1.6x \geq 38.4-12.8$

$1.6x \geq 25.6$

Divide both sides by 1.6

$\frac{1.6x}{1.6}\geq \frac{25.6}{1.6}$

∴ $x \geq 16$

Let the graphs be A, B and C from left to right

The first graph (A) shows $x\leq -8$ and this matches the 2nd inequality

The second graph (B) shows $x \geq 16$ and this matches the 3rd inequality

The third graph (C) shows $x > 9$ and this matches the 1st inequality

Hence, the correct solution graph to the inequalities are

$4(9x-18)>3(8x+12)$ → C

$-\frac{1}{3}(12x+6) \geq -2x +14$ → A

$1.6(x+8)\geq 38.4$ → B

Learn more here: brainly.com/question/17448505

8 0

3 years ago

What is true for f(x)=bx

kozerog [31]

Answer:

f'(x) = b

Step-by-step explanation:

f(x) = bx

f' (x) = d/dx (bx)

using d/dx ( a * x ) = a

f' (x) = b <-- solution.

4 0

3 years ago

Find three numbers such that their sum is 12, the sum of the first, twice the second, and three times the third is 31, and the s

Tpy6a [65]

Answer:

a = 3, b = -1, c = 10

Step-by-step explanation:

Let the three numbers be a, b and c.

Equation 1: a + b + c = 12

Equation 2: a + 2b + 3c = 31

Equation 3: 9b + c = 1

Equation 2 - Equation 1:

Equation 4: b + 2c = 19

Equation 3 times by the number 2

Equation 5: 18b + 2c = 2

Equation 5 - Equation 4

17b = -17

b = -1

Substitute into Equation 4:

2c - 1 = 19

2c = 20

c = 10

Substitute into Equation 1:

a + b + c = 12

a - 1 + 10 = 12

a = 3

8 0

3 years ago

Read 2 more answers

2/3x + 1/6 = -3/4x + 1

kupik [55]

$\frac{2}{3} x + \frac{1}{6} = -\frac{3}{4} x + 1$

First, simplify $\frac{2}{3} x$ to $\frac{2x}{3}$ / Your problem should look like: $\frac{2x}{3} + \frac{1}{6} = -\frac{3}{4} x + 1$
Second, simplify $\frac{3}{4} x$ to $\frac{3x}{4}$ / Your problem should look like: $\frac{2x}{3} + \frac{1}{6} = -\frac{3x}{4} + 1$
Third, multiply both sides by 12 (the LCM of 3 and 4) / Your problem should look like: $8x + 2 = -9x + 12$
Fourth, subtract 2 from both sides. / Your problem should look like: $8x = -9x + 12 - 2$
Fifth, simplify -9x + 12 - 2 to -9x + 10. / Your problem should look like: $8x = -9x + 10$
Sixth, add 9x to both sides. / Your problem should look like: $8x + 9x = 10$
Seventh, add 8x + 9x to get 17x. / Your problem should look like: $17x = 10$
Eighth, divide both sides by 17. / Your problem should look like: $x = \frac{10}{17}$

Answer as fraction: x = $\frac{10}{17}$
Answer as decimal: x = 0.5882

7 0

3 years ago