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3 years ago

Please help!

Mathematics

2 answers:

Svetllana [295]3 years ago

7 0

Answer:

part A: all the above

partt B: 48

part C: 4 in. x 12 in.

2 in. x 24 in.

8 in. x 6 in.

3 in. x 16 in.

Step-by-step explanation:

horsena [70]3 years ago

3 0

Answer:

Step-by-step explanation:

Part A,

99% sure this is D

Part B

It's probably about 4 feet tall.

Part C,

4 in. x 12 in.

2 in. x 24 in.

8 in. x 6 in.

3 in. x 16 in.

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.

Degger [83]

Answer:

So do you already have the answer? Also 1/20 looks right.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

4х = 32<br> А. x= 9<br> В. x= 8<br> с. x= 28<br> D. х = 128

STatiana [176]

Answer:

Step-by-step explanation:

Step 1: Divide both sides by 4.

4x/4 = 32/4

4x/4 leaves you with x by itself, and 32/4 is 8. So x = 8

3 0

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

In a secondary school there are 40 classes altogether.

fgiga [73]

So it would be 8/40. And in its simplest terms would be 1/5

7 0

3 years ago

Get the general term for the sequence being your t3 = 11 and the t20 = 244.2

marusya05 [52]

Answer:

nth term = $t_{n} = 7.639(1.2)^{n - 1}$

Step-by-step explanation:

Let us assume that the given sequence is a G.P.

Now, if the first term of the G.P. is a and the common ratio is r, then

Third term = $t_{3} = ar^{2} = 11$ .......... (1) and

20th term = $t_{20} = ar^{19} = 244.2$ ........... (2)

Now, dividing equation (2) with equation (1) we get

$\frac{ar^{19} }{ar^{2} } = \frac{244.2}{11} = 22.2$

⇒ $r^{17} = 22.2$

⇒ r = 1.2.

Hence, from equation (1) we get

a(1.2)² = 11

⇒ a = 7.639 (Approx.)

Therefore, the general term of the sequence i.e. nth term = $t_{n} = 7.639(1.2)^{n - 1}$ (Answer)

5 0

3 years ago