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3 years ago

If a jogger runs 2 miles and burned 185 calories, how many calories when he burned jogging 3 miles?

Mathematics

2 answers:

Tom [10]3 years ago

3 0

Answer:

The jogger will burn 277.5 calories

jok3333 [9.3K]3 years ago

3 0

Answer:

277.5 calories

Step-by-step explanation:

185/2 = 92.5 calories per mile

3 x 92.5 = 277.5 calories when he runs 3 miles.

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Solve the inequality -5/4(8x+4)<6-3x. Show your work.

Debora [2.8K]

Answer:

$x>-\frac{11}{7}$

Step-by-step explanation:

Step 1: Distribute

$-\frac{5}{4} (8x+4)$

$(-\frac{5}{4} *8x)+(-\frac{5}{4} *4)$

$-\frac{40x}{4}-\frac{20}{4}$

$-10x-5$

Step 2: Add 10x to both sides

$-10x-5\mathbf{+10x}$

$-5$

Step 3: Subtract 6 from both sides

$-5\mathbf{-6}$

$-11$

Step 4: Divide both sides by 7

$-11\mathbf{/7}$

$x>-\frac{11}{7}$

Answer: $x>-\frac{11}{7}$

3 0

3 years ago

Read 2 more answers

1000 grams decreased by 94%

gregori [183]

$subtract\ 94\%\ of\ 1000\ from\ 1000\\\\1000-94\%*10000=1000-0.94*1000=1000-940=60\\\\
1000\ grams\ decreased\ by\ 94\%\ is\ 60.$

3 0

3 years ago

A metal strip of 44 inches long is to be cut in 2 pieces. the smaller piece to be 12 inches shorter than the larger piece. find

Ket [755]

Answer:

Shorter Piece is 16 inches and Longer Piece is 28 Inches.

Step-by-step explanation:

Cut the 44 Inch piece in half (22+22) and subtracted 6 from one end and added six to another end which is 12 inches in total.

Hope this helps.

#FORTHEPUPPIES

3 0

3 years ago

Use the disk method or the shell method to find the volumes of the solids generated by revolving the region bounded by the graph

diamong [38]

Answer:

a) 8π

b) 8/3 π

c) 32/5 π

d) 176/15 π

Step-by-step explanation:

Given lines : y = √x, y = 2, x = 0.

a) The x-axis

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) $\int\limits^2_0 {ph} \, dy$

= $( 2\pi ) \int\limits^2_0 {y.y^2} \, dy$

∴ Vol = 8π

b) The line y = 2 ( using the shell method )

p = 2 - y

h = y^2

vol = ( 2π ) $\int\limits^2_0 {ph} \, dy$

= $( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy$

= ( 2π ) * [ 2/3 * y^3 - y^4 / 4 ] ²₀

∴ Vol = 8/3 π

c) The y-axis ( using shell method )

h = 2-y = h = 2 - √x

p = x

vol = $(2\pi ) \int\limits^4_0 {ph} \, dx$

= $(2\pi ) \int\limits^4_0 {x(2-\sqrt{x} ) } \, dx$

= ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

d) The line x = -1 (using shell method )

p = 1 + x

h = 2√x

vol = $(2\pi ) \int\limits^4_0 {ph} \, dx$

Hence vol = 176/15 π

attached below is the graphical representation of P and h

3 0

3 years ago

A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as muc

Masteriza [31]

Let x represent amount invested in the higher-yielding account.

We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be $2x$ .

We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.

$I=Prt$ , where,

I = Amount of interest,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

We are told that interest rates are 6% and 10%.

$6\%=\frac{6}{100}=0.06$

$10\%=\frac{10}{100}=0.10$

Amount of interest earned from lower-yielding account: $2x(0.06)=0.12x$ .

Amount of interest earned from higher-yielding account: $x(0.10)=0.10x$ .

$0.12x+0.10x=6600$

Let us solve for x.

$0.22x=6600$

$\frac{0.22x}{0.22}=\frac{6600}{0.22}$

$x=30,000$

Therefore, the man invested $30,000 at 10%.

Amount invested in the lower-yielding account would be $2x\Rightarrow 2(30,000)=60,000$ .

Therefore, the man invested $60,000 at 6%.

8 0

3 years ago