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Doss [256]
3 years ago
10

Mamie saved $161.25. this is 25% of the amount she needs to save how mutch money does she need to save

Mathematics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:

483.75 is needed

Step-by-step explanation:

161.25 * 4 = 645.00

645.00 - 161.25 =

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1/2 mi = 880 yd<br> O True<br> O False<br><br><br><br> Help
Black_prince [1.1K]
I don’t know I am just guessing but it’s False good luck
5 0
3 years ago
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Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)​
cricket20 [7]

xe^y+4\ln y=x^2

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}

\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}

If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

At the point (1, 1), the derivative is

\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}

4 0
3 years ago
I don't know how to answer this
scoray [572]
If you have a protractor measure the degrees in vetesis

4 0
3 years ago
Need help pls explain
Scilla [17]

Answer:

  • C. 93°

Step-by-step explanation:

WY  is the diameter.

WX and XY sum to 180° as half the circle.

<u>Central angle is equal to intercepted arc:</u>

  • mXY = m∠XVY = 87°
  • mWX = 180° - mXY = 180° - 87° = 93°

Correct choice is C

4 0
2 years ago
Read 2 more answers
Calculate the missing angle and give a reason for your answer
larisa86 [58]
Here’s the proof:

The triangle is an Isosceles Triangle. An Isosceles Triangle is a type of triangle that has 2 congruent sides. There is a Theorem called The Isosceles Triangle Theorem, in which states that if a triangle has 2 congruent sides, then it is an Isosceles Triangle, and the angles oppositely the congruent sides will be congruent. Therefore, we know that the triangle is Isosceles because it has 2 congruent sides, so the angles opposite those sides are congruent. We first need to find the angle measures.

1.) Finding the angle measures using the Triangle Sum Theorem. The Triangle Sum Theorem states that all 3 angles in a triangle must sum up to 180°. So, we can create an Algebraic equation to solve and figure out what the angle measures are of the two angles at the bottom of the triangle because they are opposite to the congruent sides.
Equation:
26+2x=180. We know that one angle measure is 26°, and we know the other 2 are angles are congruent, so we will call them X, and there are two of them, so we call them 2x.
26-26+2x=180-26
2x=154
2x/2=154/2
x=77.
Therefore, the angles measure 77°. To check that we can just plug them back into our equation:
26+2(77)=180.
180=180.
The angle measures are correct.

2.) Since we know the angle measures of the triangle, we can use the Alternate Exterior Angles Theorem, which states: if a transversal is intersected by two parallel lines, then the alternating exterior angles are congruent. We can see in the picture that the lane with the triangle on top and trapezoid at the bottom is a transversal (transversal=line intersected by two parallel lines). We can also see that the two lines with arrows are parallel because they are marked parallel with arrows. Therefore, we have a transversal intersected by two parallel lines, so the angles outside the parallel lines and and opposite each other are congruent.

We know that the two angles at the base of the triangle are 77°. The left base angle in the triangle is outside the parallel lines, so angle n or n° must also be 77° because they are both outside the parallel lines (exterior) and alternating from each other.

Therefore, n°=77°
8 0
2 years ago
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