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3 years ago

Evaluate the expression: 7 to the -5th power multiplied by 7 to the 3rd power.

Mathematics

1 answer:

Lelu [443]3 years ago

6 0

When the bases are the same you add the exponents

-5+3 ==-2
7^-2 or. 1/49

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Which is the equation of the line shown in the graph?

elixir [45]

I think the right answer is A

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3 years ago

Read 2 more answers

Bill is in a class of 15 boys and 35 girls 40 percent of the students in the class take the bus to school how many students do n

GREYUIT [131]

the gender of the kids does not matter so it can be discarded.

you add up how many boys and girls in the class to make a total of 50 kids

your 100% is 50

meaning 40% of 50 is 20

so if 20 kids took the bus to school then 30 did not take the bus.

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3 years ago

Solve the triangle.

saw5 [17]

Answer:

Step-by-step explanation:

because <u>c</u> is shorter than<u> a</u> we know C is a smaller angle than A

so the first choice is out.

use the law of cosines to find b

$b^{2}$ = $a^{2} +c^{2} -2*a*c*cos(B)$

$b^{2}$ =1681+400-1326.787871

$b^{2}$ = 754.21219

b= $\sqrt{754.21219}$

b= 27.46292

so the last choice looks good

5 0

3 years ago

Determine which of these relations are transitive. The variables x, y, x', y represent integers A.x~y if and only if x +y is pos

Veronika [31]

Answer:

A, B, C,D, F

Step-by-step explanation:

The explanation for why each option is transitive and option E is not transitive is explained in the attachment

6 0

4 years ago

1011+111 in binary form

Ann [662]

Answer:

10010

Step-by-step explanation:

$1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$111=1(2)^2+1(2)^1+1(2)^0$

So $1011+111$ gives us:

$1(2)^3+0(2)^2+1(2)^1+1(2)^0$

$+$

$1(2)^2+1(2)^1+1(2)^0$

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

Combine like terms:

$1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0$

$1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0$

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace $2^0$ with 1 and $2$ with $(2)^1$ :

$1(2)^3+1(2)^2+(2)^2+(2)^1(1)$

I want a $2^0$ number:

$1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0$

Combine like terms:

$1(2)^3+2(2)^2+1(2)^1+0(2)^0$

$2(2)^2=2^3$ :

$1(2)^3+2^3+1(2)^1+0(2)^0$

Combine like terms:

$2(2)^3+1(2)^1+0(2)^0$

We can rewrite the first term by law of exponents:

$2^4+1(2)^1+0(2)^0$

$1(2)^4+1(2)^1+0(2)^0$

So the binary form is:

$10010$

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

1 0 1 1

+ 1 1 1

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

(1) (1) (1)

1 0 1 1

+ 1 1 1

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

1 0 0 1 0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0

3 years ago