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loris [4]
2 years ago
7

Given the figure below, find the values of x and z

Mathematics
1 answer:
Dafna1 [17]2 years ago
7 0

Answer:

my bad it was a mistake .

Step-by-step explanation:

You might be interested in
Apply the square root principle to solve (x-5)^2-40=0
Nitella [24]

Answer: {5 ± 2√10, 5 - 2√10}

Step-by-step explanation: First isolate the binomial squared by adding 40 to both sides to get (x - 5)² = 40.

Next, square root both sides to get x - 5 = ± √40.

Notice that root of 40 can be broken down to 2√10.

So we have x - 5 = ± 2√10.

To get <em>x</em> by itself, add 5 to both sides to get x = 5 ± 2√10.

So our answer is just {5 ± 2√10, 5 - 2√10}.

As a matter of form, the number will always come before the

radical term in your answer to these types of problems.

In other words, we use 5 ± 2√10 instead of ± 2√10 + 5.

4 0
3 years ago
Sound travels through sea water at a speed of about 1500 meters per second. At this rate, how far will sound travel in 2 minutes
Mariulka [41]

Answer:

180,000 m

Step-by-step explanation:

There are 60 seconds in a minute, so:

2 min × 60 s/min = 120 s

Distance = rate × time

d = 1500 m/s × 120 s

d = 180,000 m

8 0
3 years ago
Simplify the following expression:
nalin [4]
-2/2 I think probably
3 0
3 years ago
Tom has a solid, wooden block that is a rectangular prism with non-square bases. He saws the prism parallel to the base. What sh
Katyanochek1 [597]

Answer:

A) Rectangle

Step-by-step explanation:

Hope this helps :D

6 0
3 years ago
Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0
Leokris [45]

Substitute v(x)=-2x+y(x), so that \dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}. Then the ODE is equivalent to

\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7

which is separable as

\dfrac{\mathrm dv}{v^2-9}=\mathrm dx

Split the left side into partial fractions,

\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)

so that integrating both sides is trivial and we get

\dfrac{\ln|v-3|-\ln|v+3|}6=x+C

\ln\left|\dfrac{v-3}{v+3}\right|=6x+C

\dfrac{v-3}{v+3}=Ce^{6x}

\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}

\dfrac6{v+3}=1-Ce^{6x}

v=\dfrac6{1-Ce^{6x}}-3

-2x+y=\dfrac6{1-Ce^{6x}}-3

y=2x+\dfrac6{1-Ce^{6x}}-3

Given the initial condition y(0)=0, we find

0=\dfrac6{1-C}-3\implies C=-1

so that the ODE has the particular solution,

\boxed{y=2x+\dfrac6{1+e^{6x}}-3}

5 0
3 years ago
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