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2 years ago

PLZZZ HELP IF YOU ANSWER THIS JUST TO GET POINTS I WILL REPORT YOU PLZZZ I NEED HELP

Mathematics

2 answers:

Ad libitum [116K]2 years ago

8 0

Answer:

1372 inches

Step-by-step explanation:

980/5=196

196*7=1372

faltersainse [42]2 years ago

3 0

The answer is 196 inches. Therefore the ant can travel 196 inches per minute.

An ant can travel at a constant speed of 980 inches every 5 minutes.

What you should do is find the distance traveled from this ant by in one minute so that we can get the accuracy from just dividing the total distance that was from this place to place distance.

Traveled distance from an ant in 1 min=980/5

Distance that was traveled by an ant is in 1 minute so 196.

Please give brainiest

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4.5 - 17 - X 8 I need help! 10 pts

sesenic [268]

Answer:

x = -1.5625

Step-by-step explanation:

Solve for x:

-8 x - 12.5 = 0

Add 12.5` to both sides:

(12.5 - 12.5) - 8 x = 12.5

12.5 - 12.5 = 0:

-8 x = 12.5

Divide both sides of -8 x = 12.5 by -8:

(-8 x)/(-8) = 12.5/(-8)

(-8)/(-8) = 1:

x = 12.5/(-8)

12.5/(-8) = -1.5625:

Answer: x = -1.5625

7 0

2 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$

4 0

3 years ago

In a group of 21 students, 8 students take Art courses, 7 students take Music courses. Two students take both courses. A student

Blababa [14]

Since two people take both courses, we can subtract two from 7, 8, and 21, giving us 5, 6, and 19.

Add 5 and 6 together to get 11. Since eleven students are in those two groups, we can answer this. The answer will be 11/19.

3 0

2 years ago

Two angles have a common vertex but no common side. How can you tell whether the angles are vertical angles?

pentagon [3]

Vertical angles is an X but that's how you tell if they're vertical angles

8 0

2 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

6 0

1 year ago