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1 year ago

Find the length of EH

Mathematics

1 answer:

san4es73 [151]1 year ago

3 0

Answer:

4 (5/6)

Step-by-step explanation:

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al dividir A entre 23 se obtuvo 247 de cociente y el residuo fue el maximo. hallar A a)5242 b)5436 c)5703 d)5332 solo esas 4 alt

yuradex [85]

Aplicando la división, hay que el valor de A es dado por:

c) 5703

Una división es representada por:

$\frac{a}{b} = c + \frac{r}{b}$

En que:

c es el cociente.

r es el residuo.

En este problema, $b = 23$ , y también:

Cociente de 247, o sea, $c = 247$ .

Residuo maximo, que es 22, o sea, $r = 22$ .

$\frac{a}{b} = c + \frac{r}{b}$

$\frac{A}{23} = 247 + \frac{22}{23}$

$A = 23(247) + 22 = 5703$

El valor de A es dado por:

c) 5703

Un problema similar, que tambíen envuelve división, es dado en brainly.com/question/14270856

8 0

2 years ago

What is the length of segment TR with T(-4,5) and R(2, -1)

Brilliant_brown [7]

Answer:

The length of segment TR with T(-4,5) and R(2, -1) is 8.49

Step-by-step explanation:

We need to find the length of segment TR with T(-4,5) and R(2, -1)

The length of segment can be found using distance formula.

The distance formula is: $Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Putting values and finding length

We have: $x_1=-4, y_1=5, x_2=2, y_2=-1$

$Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance = \sqrt{(2-(-4))^2+(-1-5)^2}\\Distance = \sqrt{(2+4)^2+(-1-5)^2}\\Distance = \sqrt{(6)^2+(-6)^2}\\Distance = \sqrt{36+36}\\Distance = \sqrt{72}\\Distance=8.49$

We found Distance = 8.49

So, The length of segment TR with T(-4,5) and R(2, -1) is 8.49.

5 0

3 years ago

I don't understand exponents

Firlakuza [10]

You must apply the law of exponets.

You must apply only at the same letter. X or Y

$\frac{x^{m}}{x^{n}} = x^{m-n}
$

$\frac{x^{3}}{x^{6}} = x^{3-6} =x^{-3}$

$\frac{y^{3}}{y^{-6}} = y^{(3-(-6))} =y^{3+6}=y^{9}$

$\frac{20x^{3}y^{3}}{32x^{6}y^{-6}} = \frac{20x^{(3-6)}y^{(3-(-6))}}{32}=\frac{20x^{(-3)}y^{(3+6)}}{32}=\frac{20x^{(-3)}y^{(9)}}{32}
$

But
$x^{-n} = \frac{1}{x^{n}}$
Then

$\frac{20x^{3}y^{3}}{32x^{6}y^{-6}} = \frac{20x^{(-3)}y^{(9)}}{32}=\frac{20y^{9}}{32x^{3}}=\frac{10y^{9}}{16x^{3}}=\frac{5y^{9}}{8x^{3}}$

5 0

3 years ago

Please help me with this. 20 points.

antoniya [11.8K]

Answer:A,C

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A man traveled 24 miles in 20 minutes and 18 miles in 15 minutes. Assume that the distance varies linearly with time. How far wi

Norma-Jean [14]

Step-by-step explanation:

First we need to get his speed:

speed = distance / time

If he travelled 24 miles in 20 mintues, his speed is 24 / 20 = 1.2 miles per minute

If he travelled 18 miles in 15 minutes, his speed is 18 / 15 = 1.2 miles per minute

Those two should give us the same answer since it says that the distance varies linear with time, which means that his speed is constant.

Then after 30 minutes he will have traveled: 1.2 x 30 = 36 miles

please mark me as brainliest

4 0

2 years ago