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2 years ago

Solve similar triangles can someone please answer please

Mathematics

1 answer:

Dahasolnce [82]2 years ago

6 0

Answer:

Step-by-step explanation:

ΔADE & ΔABC are similar triangles.

$\frac{AD}{AB}=\frac{DE}{BC}\\\\\frac{12}{4}=\frac{12}{x}\\$

Cross multiply,

12*x = 12*4

$x= \frac{12*4}{12}\\$

x = 4

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Write the following fraction as a decimal. 2<img src="https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B10%7D" id="TexFormula1" title

vfiekz [6]

Answer:

2/10 or 1/2 = .5 you would divide 2 by 2 and 10 divided by 2 (whatever you multiply or divide by on the top you must do on the bottom)

2/10=.50

3/100=.03 (divide 3 by 10=.0

5 0

2 years ago

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Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =

irga5000 [103]

Answer: The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

8 0

3 years ago

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Napa and Newton are 24 miles apart. On a map, the two cities are 3 inches apart. What is the map scale?

Andreas93 [3]

Answer:

one inch on the map is 8 miles

Step-by-step explanation:

divide 24 by 3

5 0

3 years ago

I need answer Immediately pls!!!!!!

Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

$n(U)=27$

$n(A)=7$

$n(B)=18$

$n(A'\cap B')=7$

We know that,

$n(A\cup B)=n(U)-n(A'\cap B')$

$n(A\cup B)=27-7$

$n(A\cup B)=20$

Now,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=7+18-n(A\cap B)$

$n(A\cap B)=7+18-20$

$n(A\cap B)=25-20$

$n(A\cap B)=5$

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

$\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}$

$\text{Probability}=\dfrac{5}{27}$

Therefore, the required probability is $\dfrac{5}{27}$ .

3 0

3 years ago

15/40 in lowest term I need it so bad now

Vesnalui [34]

In order to reduce ANY fraction to lowest terms, find any common factors
of the numerator and denominator, and divide them both by it. If they still
have a common factor, then divide them by it again. Eventually, they won't
have any common factor except ' 1 ', and then you'll know that the fraction is
in lowest terms.

Do 15 and 40 have any common factors ?
Let's see . . .

The factors of 15 are 1, 3, 5, and 15 .
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40 .

Ah hah ! Do you see that ' 5 ' on both lists ? That's a common factor.

So 15/40 is NOT in lowest terms.

Divide the numerator and denominator both by 5 :

15 / 40 = 3 / 8

3 and 8 don't have any common factor except ' 1 '.
So 3/8 is the same number as 15/40, but in lowest terms.

5 0

3 years ago

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