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3 years ago

What is x given triangleABC ~triangleDBE?

Mathematics

1 answer:

denis-greek [22]3 years ago

7 0

x = 37.5 (or) $\frac{75}{2}$

Solution:

Given $\triangle A B C \sim \triangle D B E$ .

Let us take BE = x and BC = 25 + x.

To determine the value of x:

If two triangles are similar then the corresponding angles are congruent and the corresponding sides are in proportion.

$$\frac{AC}{DE}=\frac{B C}{B E}$

$$\frac{50}{30} =\frac{25+x}{x}$

Do cross multiplication, we get

$50x=30(25+x)$

$50x=750+30x$

Subtract 30x from both sides of the equation.

$20 x=750$

Divide by 20 on both sides of the equation, we get

x = 37.5 (or) $\frac{75}{2}$

Hence the value of x is 37.5 or $\frac{75}{2}$ .

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HELP!!!!!!!!! ME!!!!!!!!! HELP!!!!

Gala2k [10]

Correct answer is the last one.
Function because for every x only one value of y,
not one-to-one because, 2 values of x have the same value of y.

7 0

4 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

3 years ago

a. Determine whether the Mean Value Theorem applies to the function f (x )equals ln 15 xf(x)=ln15x on the given interval [1 comm

ivolga24 [154]

Answer:

(a)

The function f is continuous at [1,e] and differentiable at (1,e), therefore

the mean value theorem applies to the function.

(b)

$c = e-1 \\$ = 1.71828

Step-by-step explanation:

(a)

The function f is continuous at [1,e] and differentiable at (1,e), therefore

the mean value theorem applies to the function.

(b)

You are looking for a point $c$ such that

$\frac{1}{c} = \frac{\ln(15e)-\ln(15*1)}{e-1} = \frac{\ln(15e/15)}{e-1} = \frac{\ln(e)}{e-1} = \frac{1}{e-1}$

You have to solve for $c$ and get that

$c = e-1 \\$ = 1.71828

5 0

3 years ago

Solve this system of equations: 5x + y = 7 20x + 2 = y

xz_007 [3.2K]

Answer:

x = 1/5, y = 6

Step-by-step explanation:

5x + y = 7

y = 7 - 5x

Set equations equal to each other to eliminate y

7 - 5x = 20x + 2

7 = 25x + 2

5 = 25x

x = 1/5

y = 7 - 5(1/5)

y = 7 - 1

y = 6

7 0

3 years ago

How many digits are written when 1000 to the power of 2015 is expressed as a numeral?

PtichkaEL [24]

Answer: There are 6046 digits will be written on expansion upto 2015th power.

Step-by-step explanation:

Since we have given that

$1000^{2015}$

We have to find the number of digits ,

So,

$1000^{2015}\\\\=(10^3)^{2015}\\\\=10^{6045}$

Since we know that

$10^n\text{ has n+1 number of digits }$

so,

$10^{6045}\text{ has 6046 number of digits }$

Hence, there are 6046 digits will be written on expansion upto 2015th power.

6 0

3 years ago