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3241004551 [841]

1 year ago

9

The equation of the graphed line is x+2y=5. What is the x-intercept of the graph?у65432-12345 6-6-5-4-3-2-11-2on a WN-6O22.55Sav

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1 answer:

Norma-Jean [14]1 year ago

4 0

The x-intercept is the value of x where the graph intersects the x-axis.

Looking at the graph, we can see that the line intersects the x-axis at the value x = 5.

Therefore the correct option is the third one.

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The sum of one and the product 4 and a number z

Dovator [93]

Answer: 5z

Step-by-step explanation:

1+4 * z= 5z

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3 years ago

i need help!!! Match each power of ten with the appropriate number. 1. 100 2. 102 3. 10-3 4. 10-2 a. 0.01 b. 0.001 c. 1 d. 100

Masja [62]

1. 100 -> 10²

2. 102 -> 10² + 2

3. 10^-3 -> 1/(1000)

4. 10^-2 -> 1/(100)

a. 0.01 -> 10^-2

b. 0.001 -> 10^-3

c. 1 -> 1

d. 100 -> 10^2

hope this helps

5 0

3 years ago

What is the solution to this equation?<br> (1/4)x+1 =32<br> A. -7/2<br> B. -2<br> C. 3/2<br> D. 2

ANEK [815]

(1/4)x+1=32
(1/4)x=31
X=31*4
X=124

7 0

2 years ago

This week, Jolene withdrew $225 from her savings account. She made a deposit of $125 then bought $75.35 worth of groceries using

Ulleksa [173]

Answer:

675.7 dollars

Step-by-step explanation:

1) 475.35 + 75.35 + 125 = 675.7

6 0

3 years ago

Please I need help with differential equation. Thank you

Inga [223]

1. I suppose the ODE is supposed to be

$\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)$

Solving for $\dfrac{\mathrm dy}{\mathrm dt}$ gives

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}$

which is undefined when $t=\pm1$ . The interval of validity depends on what your initial value is. In this case, it's $t=-\dfrac12$ , so the largest interval on which a solution can exist is $-1\le t\le1$ .

2. Separating the variables gives

$\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}$

Integrate both sides. On the left, we have

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}$

where we substituted $z=y^{1/2}$ - or $z^2=y$ - and $2z\,\mathrm dz=\mathrm dy$ - or $\mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}$ .

$\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)$

On the right, we have

$\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)$

$\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C$

So

$2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C$

$\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C$

$y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}$

$y^{1/2}=C(1-t^2)^{1/4}-1$

I'll leave the solution in this form for now to make solving for $C$ easier. Given that $y\left(-\dfrac12\right)=1$ , we get

$1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1$

$2=C\left(\dfrac54\right)^{1/4}$

$C=2\left(\dfrac45\right)^{1/4}$

and so our solution is

$\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}$

3 0

3 years ago