All categories

3 years ago

I need help , if yk the answer help fr!!!

Mathematics

1 answer:

Feliz [49]3 years ago

4 0

Answer:

The measure of the angle JKG is:

m∠JKG = 56°

Step-by-step explanation:

<u>Given</u>

m∠JKG = 76-2x

m∠FHK = 6x-4

J is a midpoint of the segment FG and K is a midpoint of the segment GH.

<u>To determine</u>

m∠JKG = ?

Given that J is a midpoint of the segment FG and K is a midpoint of the segment GH. Thus, making two similar triangles, ΔJGK and ΔFGH

We know that two triangles are similar if the only difference is size. So, the angles remain the same.

so m∠JKG and m∠FHK are equal.

i.e.

m∠JKG = m∠FHK

substitute m∠JKG = 76-2x and m∠FHK = 6x-4

76-2x = 6x-4

6x+2x = 76 + 4

8x = 80

divide both sides by 8

8x/8 = 80/8

x = 10

Therefore, the value of x = 10

m∠JKG = 76-2x

substitute x = 10

m∠JKG = 76 - 2(10)

= 76 - 20

= 56°

Therefore, measure of the angle JKG is:

m∠JKG = 56°

You might be interested in

What times 2 is equal to 20?

alina1380 [7]

X×2=10 X=10÷2 X=5 Proof: 5×2=10

7 0

3 years ago

Read 2 more answers

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

4 years ago

What is the volume of this rectangular prism?

goblinko [34]

$5 \times 5 \times 10 = 250$

5 0

3 years ago

Hi Does anyone know this?

Contact [7]

The answer is going to be s. The proportions are (short leg/long leg) , and the appropriate long leg for h is s.

8 0

4 years ago

For breakfast, Randy had two Egg

elixir [45]

Since the only difference in the two meals is one egg mcmuffin, then the difference in the calorie count has to be due to the mcmuffin. So

750 - 450 = 300 calories for an egg mcmuffin.

Jack's egg mcmuffin and hashbrowns totaled 450 calories. Since we know that an egg mcmuffin is 300 calories, then the hashbrowns have to be

450 - 300 = 150 calories

8 0

4 years ago