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3 years ago

8

Which group of interior angle measurements can form a triangle? Check all that apply.

2 answers:

Juliette [100K]3 years ago

4 0

Answer:

110°, 71°, 62°

110°, 30°, 40°

2 years ago

yeah only one of your answers is there

xz_007 [3.2K]3 years ago

3 0

Answer:

123,35,22

47,71,62

110,30,40

Step-by-step explanation:

Sum of all angles of a triangle = 180

123 + 35 + 22 = 180

47 + 71 + 62 = 180

110 + 30 + 40 = 180

2 years ago

and only one of your answers is there also

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Two bicycle wheels are shown below which is closest to the circumference in inches of the larger wheel.

olganol [36]

We estimate the diameter of the larger wheel to be 24 to 25 inches.

The formula for the Circ. of a circle in terms of the diameter is C = pi*d.

Here, the Circ. of the larger wheel is approx C = pi(24 inches) = 24*pi inches,
or

24(3.14) inches = 75.36 inches. You must now choose from the four answer choices. Which one is closes to 75.36 inches?

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3 years ago

Select the correct answer. Which expression is equivalent to 8x^2^3 sqrt 375x + 2^3 sqrt 3x^7, if x=0?

natima [27]

Answer:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}$

Step-by-step explanation:

The question is poorly formatted. The original question is:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7$

We have:

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7$

Open bracket

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

Express 8 as 2^3

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^{3* \frac{2}{3}} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^2 *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}$

Express 2^3 as 8

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{75x^3} + 8\sqrt{ 3x^7}$

Expand each exponent

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2 *3x} + 8\sqrt{ 3x * x^6}$

Split

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 *5x *\sqrt{3x} + 8\sqrt{3x} * x^3$

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =20x^3\sqrt{3x} + 8x^3\sqrt{3x}$

Factorize

$(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =28x^3\sqrt{3x}$

4 0

3 years ago

Read 2 more answers

Using 3, 9, 2 and 8 how do you get 38?

leonid [27]

Answer:38

Step-by-step explanation:

you put 3 in front of 8 and you get 38

6 0

3 years ago

What is the solution for x in the equation? 16x − 4 + 5x = -67 A.x=-3 B.x=3 C.x=1/3 D. x=-1/3

lozanna [386]

Answer:

A. X=-3

Step-by-step explanation:

16x-4+5x=-67

21x=-67+4

21x=-63

Divide through by 21

X=-3

8 0

3 years ago

If q = r and r = s, then q = s.

elena-14-01-66 [18.8K]

Answer:

c) transitive property

3 0

3 years ago