Jazmine needs $30,000 to pay off a loan at the end of 5 years. How much must she deposit monthly into

Please help quickly!!!!!!!!!!The area of the regular pentagon is 6.9 cm2. What is the perimeter? 2 cm 5 cm 10 cm 20 cm

pishuonlain [190]

10
random words to fill up 20 character minimum for answering questions :P

7 0

2 years ago

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A bike wheel is 26 inches in diameter. What is the bike wheel's diameter in millimeters (1 inch = 25.4 millimeters)?

ruslelena [56]

Answer:

Diameter of wheel in millimetres is 660.4

Step-by-step explanation:

Diameter of wheel in inches = 26

given

1 inch = 25.4 millimeters

multiplying RHS and LHS by 26

26*1 inch = 26*25.4 millimeters

=>26 inch = 660.4 mm.

Thus, diameter of wheel in millimetres is 660.4

3 0

2 years ago

Jackie bought a 3-pack of tubes of glue. She used 0.27 ounce of the glue from the 3-pack

brilliants [131]

Answer:

0.54 ounces in each tube

Step-by-step explanation:

Add the amount she used so we can add it in. 1.35+0.27= 1.62. Then divide it by the number of tube and since it was a 3 pack you divide it by 3. 1.62/3=0.54 so there is 0.54 ounces in each tube.

5 0

2 years ago

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A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the plate, including t

vredina [299]

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

$\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2$

$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$ . But this point does not fall in the region $x^2+y^2\le1$ . There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of $x^2+y^2\le1$ by

$x=\cos u$

$y=\sin u$

with $0\le u$ . Then $t(x,y)$ can be considered a function of $u$ alone:

$t(x,y)=t(\cos u,\sin u)=T(u)$

$T(u)=\cos^2u+3\sin^2u+13\cos u$

$T(u)=3+13\cos u-2\cos^2u$

$T(u)$ has critical points where $T'(u)=0$ :

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

$T(0)=14$

$T(\pi)=-12$

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0

1 year ago

"On a map, 0.25 inches represents 1 mile. What is the area of a rectangle on the map that is 3.75 inches long and 2.25 inches wi

Vika [28.1K]

0.25 inches represents 1 mile

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Find 1 inch:
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1 inch represents 1 ÷ 0.25 miles
1 inch represents 4 miles

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Find 3.75 inches:
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1 inch represents 4 miles
3.75 inches represent 4 x 3.75 miles
3.75 inches represent 15 miles

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Find 2.25 inches:
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1 inch represents 4 miles
2.25 inches represent 4 x 2.25 miles
2.25 inches represent 9 miles

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Find Area:
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Area = Length x Width
Area = 15 x 9
Area = 135 miles²

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Answer: 135 miles²
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6 0

2 years ago

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