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zubka84 [21]
3 years ago
6

Part A

Mathematics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

a) The y-intercept is 160. It represents the initial distance Jayden was from his home.

b) The x-intercept is 4. It represents the amount of time taken for Jayden to reach his home.

Step-by-step explanation:

The y-intercept can be found by looking at the value of y when x=0. When x=0, y= 160. Since the x-axis is the amount of time in hours, x=0 would mean the initial time. The x-axis is the number of miles from Jayden's home thus, Jayden was 160 miles away from his home when x=0 hours.

The x-intercept can be found by looking at the x value when y=0. When y=0, x=4. This means that it took Jayden 4 hours to reach home or 4 hours to travel 160miles to his house.

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Drag expressions to show the steps to create an equivalent expression using only positive exponents. Let g and h be two nonzero
Liula [17]

Equivalent expressions are expressions that have the same value.

The equivalent expression of g^{-3}h is: \frac{h}{g^3}

The expression is given as;

g^{-3}h

Apply law of indices to change the exponents

g^{-3}h = \frac{1}{g^3}h

Express as products

g^{-3}h = \frac{1}{g^3} \times h

Rewrite as

g^{-3}h = \frac{h}{g^3}

Hence, the equivalent expression is: \frac{h}{g^3}

Read more about equivalent expressions at:

brainly.com/question/24242989

3 0
3 years ago
I need help plz no links
Mkey [24]

Answer:

look at the picture I sent

3 0
3 years ago
Help! I already tried -6.6/20
sattari [20]

Answer: 0.33 or 1.32 or 6.6%

Step-by-step explanation:

4 0
4 years ago
Solve this problem by using variation of parameters method.<br> y''-y=coshx.
Gekata [30.6K]
y''-y=0\implies r^2-1=0\implies r=\pm1
\implies y_c=C_1e^x+C_2e^{-x}={C^*}_1\underbrace{\cosh x}_{y_1}+{C^*}_2\underbrace{\sinh x}_{y_2}

For the nonhomogeneous ODE

y''-y=\underbrace{\cosh x}_{f(x)}

we're looking for a particular solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the two fundamental solutions.

We have

W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1

so we're left with

u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x
u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x

so that the particular solution is

y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x
y_p=-\dfrac12\cosh x+\dfrac12x\sinh x

As y_1 already accounts for the \cosh x term in y_p, we're left with the general solution

y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x
3 0
3 years ago
Please help !!!! I’m desperate people.
Julli [10]

Answer:

I believe the answers are a and e.

Step-by-step explanation:

Let me know if you want an explanation.

Hope this helps!!!

8 0
2 years ago
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