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2 years ago

Round the number to the indicated place. $13.59562 to cents

Mathematics

1 answer:

Aleks04 [339]2 years ago

7 0

Answer:

look below

Step-by-step explanation:

Lets round the 0.59562 first

0.59562 rounded to the nearest hundreth (because thats how cents go) rounds up to .60

there is 100 cents in a dollar

there is 13 dollars

13 * 100 is 1300

1300 + 60

1360 cents

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The area of a circular sun spot is growing at a rate of 1,200 km2/s.

Tpy6a [65]

A)

$\bf \textit{area of a circle}\\\\
A=\pi r^2
\\\\\\
\cfrac{dA}{dt}=\pi \cdot 2r\cfrac{dr}{dt}\implies \cfrac{dA}{dt}=2\pi r\cfrac{dr}{dt}\implies \cfrac{\frac{dA}{dt}}{2\pi r}=\cfrac{dr}{dt}\\\\
-----------------------------\\\\
\left. \cfrac{dr}{dt} \right|_{r=3000}\implies \cfrac{1200}{2\pi \cdot 3000}=\cfrac{dr}{dt}\\\\$

B)

$\bf A=\pi r^2\qquad A=490000\implies 490000=\pi r^2\implies \sqrt{\cfrac{490000}{\pi }}=r
\\\\\\
\left. \cfrac{dr}{dt} \right|_{r=\sqrt{\frac{490000}{\pi }}}\implies \cfrac{1200}{2\pi \cdot \sqrt{\frac{490000}{\pi }}}=\cfrac{dr}{dt}
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8 0

3 years ago

For which fraction pairs would it be useful to use the benchmarks 0, One-half, and 1 to compare? Check all that apply.

koban [17]

Answer:

Step-by-step explanation:

thank my answer

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8 0

3 years ago

Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has

Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Solving

∆ABC A(11, 6), B(5, 6), and C(5, 17)

AB = 6 units BC = 11 units AC = 12.53 units
∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)
XY = 7.14 units YZ = 18.79 units XZ = 11.66 units

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

MN = 6 units NO = 11 units MO = 12.53 units
∆JKL J(17, -2), K(12, -2), and L(12, 7).
JK = 5 units KL = 9 units JL = 10.30 units
∆PQR P(12, 3), Q(12, -2), and R(3, -2)
PQ = 5 units QR = 9 units PR = 10.30 units
Therefore
we have the ∆ABC and the ∆MNO 
with all three sides equal ---------> are congruent
we have the ∆JKL and the ∆PQR
with all three sides equal ---------> are congruent

let's check

Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

1) If ∆MNO ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC

then ∆MNO ≅ ∆ABC

a) Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

M(-9, -4)-----------------> M1(-9,4)

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

b) Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

∆M1N1O1 M1(-9, 4), N1(-3, 4), and O1(-3, 15).

M1(-9, -4)-----------------> M2(9,4)

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

c) Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

∆M2N2O2 M2(9,4), N2(3,4), and O2(3, 15).

M2(9, 4)-----------------> M3(11,6)=A

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

∆ABC A(11, 6), B(5, 6), and C(5, 17)

∆MNO reflection------- > ∆M1N1O1 reflection---- > ∆M2N2O2 translation -- --> ∆M3N3O3

The ∆M3N3O3=∆ABC

Therefore ∆MNO ≅ ∆ABC - > check list

2) If ∆JKL -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR

then ∆JKL ≅ ∆PQR

d) Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

∆JKL J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)-----------------> J1(2,17)

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

e) translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).

J1(2, 17)-----------------> J2(12,3)=P

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

∆PQR P(12, 3), Q(12, -2), and R(3, -2)

∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR

Therefore ∆JKL ≅ ∆PQR - > check list

6 0

3 years ago

What is 39.79949748 rounded to the nearest hundredth?

Rufina [12.5K]

Answer:

$Number = 39.80$

Step-by-step explanation:

Given

$Number = 39.79949748$

Required

Approximate (to the nearest 100th)

This means that, we approximate at the second digit after the decimal.

So:

i.e,

Number = 39.79 [Begin approximation] 949748

The first digit after [Begin approximation] is then approximated using the following rule:

$0 - 4 \approx 0$

$5 - 9 \approx 1\\$

Since 9 falls in $5 - 9 \approx 1\\$ category, the number becomes:

$Number = 39.[79+1]$

$Number = 39.80$

3 0

2 years ago

Three boxes have a total weight of 640 pounds. Box A weights twice as much as Box B. Box C weight 30 pounds more than Box A. How

Zarrin [17]

Answer:

A: 244 pounds

B: 122 pounds

C: 274 pounds

Step-by-step explanation:

We have A+B+C = 640; A=2B; C=A+30. Substituting the last into the first gives ...

A + B + (A +30) = 640

2A +B = 610 . . . . . . . . . . . . subtract 30

Substituting the second into this equation gives ...

2(2B) +B = 610

B = 610/5 = 122 . . . . . divide by 5

A = 2B = 244 . . . . . . . .find A from B

C = A+30 = 274 . . . . . find C from A

Box A weighs 244 pounds; box B weighs 122 pounds; box C weighs 274 pounds.

4 0

3 years ago