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2 years ago

15

Which is the slope of the line that passes through the points (3,17)(3,17) and (7,25)(7,25)?

1 answer:

PilotLPTM [1.2K]2 years ago

4 0

The slope is 2, because you subtract the points 3,17 from 7,25 to get 8/4/ Simplify that into 2.

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Please answer soooooooooooooooon :|

madam [21]

Answer: 200 square units

Step-by-step explanation:

first you think what 60/3 is and you get 20

then you multiply 20x20 and get 400

then you divide by 2 because to find the area of a triangle you use the formula (lxh)/2

4 0

2 years ago

If the area of a square plot of<br> land is 9 miles squared, what is<br> the perimeter of this land?

tatyana61 [14]

Answer:

$\huge{ \fbox{ \sf{12 \: miles}}}$

Step-by-step explanation:

Given,

Area of a square plot = 9 miles

<u>First</u><u>,</u><u> </u><u>finding</u><u> </u><u>the</u><u> </u><u>length</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>land</u>

$\boxed{ \sf{Area \: of \: a \: square \: = \: {length}^{2} }}$

$\mapsto{ \sf{9 = {length}^{2} }}$

$\mapsto{ \sf{ {length}^{2} = 9}}$

$\mapsto{ \sf{ \sqrt{ {(length)}^{2} }} = \sqrt{9} }$

$\mapsto{ \sf{length = 3 \: miles}}$

<u>Now</u><u> </u><u>,</u><u> </u><u>finding</u><u> </u><u>the</u><u> </u><u>perimeter</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>land</u> :

$\boxed{ \sf{perimeter \: of \: a \: square = 4 \: * length}}$

$\mapsto{ \sf{perimeter \: = \: 4 \: * \: 3}}$

$\mapsto{ \sf{perimeter \: = 12 \: miles}}$

The perimeter of the land = 12 miles

Hope I helped!

Best regards! :D

~ $\text{TheAnimeGirl}$

3 0

2 years ago

Which is the correct cofunction identity for tan0

svetoff [14.1K]

Answer:

E. cot(90° -θ)

Step-by-step explanation:

The "cofunction" is named by adding or deleting the "co" in front of the function name. The cofunction for tangent is <em>cotangent</em>.

The tangent of an angle is the cotangent of the complement of the angle:

tan(θ) = cot(90° -θ)

5 0

1 year ago

Read 2 more answers

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

2 years ago

Consider this Venn diagram of the prime factors of 210 and 330. Which is the least common multiple of these two numbers?

Mrrafil [7]

The least common multiple of these two numbers is 2310. Hope this helps

5 0

2 years ago