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Ulleksa [173]
3 years ago
12

So far, Keith has burned 604.3 calories. He wants to burn a total of 700 calories. How many more calories must Keith burn?

Mathematics
1 answer:
Vanyuwa [196]3 years ago
7 0
So to do this you need to use subtraction to find it how many more calories he needs to burn 700. so you would do 700 - 604.3 to get an answer of 95.7! :)
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Find the percent of decrease from 280 to 210. Round to the nearest tenth of a percent, if necessary.
Vlada [557]
To find the decreased percentage,here's what we can do:
\frac{originl \: value - new \: value}{originl \: value}  \times 100\%

In this case,the eqaution would be:
\frac{280 - 210}{280}  \times 100\% \\  =  \frac{70}{280}  \times 100\% \\  =  \frac{1}{4}  \times 100\% \\  = 25\%
Hope it helps!
5 0
3 years ago
Trigonometric equations<br><br> 5sinx = 3sinx + square root of 2
Lera25 [3.4K]

\qquad\qquad\huge\underline{{\sf Answer}}♪

Let's solve for x ~

\qquad \sf  \dashrightarrow \:5 \sin(x)  = 3 \sin(x)   +  \sqrt{2}

\qquad \sf  \dashrightarrow \:5 \sin(x)  - 3 \sin(x)  =  \sqrt{2}

\qquad \sf  \dashrightarrow \:2 \sin(x)  =  \sqrt{2}

\qquad \sf  \dashrightarrow \: \sin(x)  =  \sqrt{2}  \div 2

\qquad \sf  \dashrightarrow \: \sin(x) =  \frac{1}{ \sqrt{2} }

\qquad \sf  \dashrightarrow \:x = 45 \degree \:  \: or \:  \:  \frac{\pi}{4}  \: rad

・ .━━━━━━━†━━━━━━━━━.・

4 0
2 years ago
6. Find the perimeter of the figure below
liraira [26]

Answer:

26 in.

Step-by-step explanation:

» <u>Concepts</u>

The perimeter for a shape is the total distance around it measured in units. The formula to find the perimeter of a trapezoid is a + b + c + d, where a & b are the bases, and c & d are the lengths of the other sides.

» <u>Solution</u>

Adding up all the sides, we get:

  • 11+7+4+4
  • 18+4+4
  • 22+4
  • 26

26 inches

3 0
2 years ago
12x - 2y = -1<br>+ 4x + 6y= -4<br>​
svp [43]

Answer:

x = -7/40 , y = -11/20

Step-by-step explanation:

Solve the following system:

{12 x - 2 y = -1 | (equation 1)

4 x + 6 y = -4 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:

{12 x - 2 y = -1 | (equation 1)

0 x+(20 y)/3 = (-11)/3 | (equation 2)

Multiply equation 2 by 3:

{12 x - 2 y = -1 | (equation 1)

0 x+20 y = -11 | (equation 2)

Divide equation 2 by 20:

{12 x - 2 y = -1 | (equation 1)

0 x+y = (-11)/20 | (equation 2)

Add 2 × (equation 2) to equation 1:

{12 x+0 y = (-21)/10 | (equation 1)

0 x+y = -11/20 | (equation 2)

Divide equation 1 by 12:

{x+0 y = (-7)/40 | (equation 1)

0 x+y = -11/20 | (equation 2)

Collect results:

Answer:  {x = -7/40 , y = -11/20

6 0
3 years ago
If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that
Ivan

Answer:

\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• \sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

•\sf\implies cos2A =\dfrac{\sqrt3}{2}

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

\sf\implies cos2A = 1 - 2sin^2A

Therefore , here substituting the value of sinA , we have ,

\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2

Simplify the whole square ,

\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}

Add the numbers in numerator ,

\sf\implies cos2A =  1-2\times \dfrac{4-2\sqrt3}{8}

Multiply it by 2 ,

\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}

Take out 2 common from the numerator ,

\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}

Simplify ,

\sf\implies cos2A =  1 -\dfrac{ 2-\sqrt3}{2}

Subtract the numbers ,

\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}

Simplify,

\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }

Hence Proved !

8 0
3 years ago
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