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3 years ago

14

After riding a bike for 20 min you burn 190 calories. If you ride the bike for 45 minutes how many calories will you burn? Use a

proportion to solve.

1 answer:

tangare [24]3 years ago

7 0

Multiple 20 times 45 and you get 900

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Plz help with number 1

Trava [24]

In an isosceles triangle, the base angles are congruent. The third angle is called the vertex angle.

Here, the vertex angle is <A.

Therefore, m<C = m<B.

m<A = 3m<B + 20

m<A + m<B + m<C = 180

3m<B + 20 + m<B + m<B = 180

5m<B + 20 = 180

5m<B = 160

m<B = 32

m<C = m<B = 32

Answer: m<C = 32 deg

4 0

3 years ago

What is QR ?<br><br> Enter your answer in the box.<br><br> units

Dimas [21]

TR = 6
QR = 2(TR)
QR = 2(6)
QR = 12

6 0

3 years ago

Read 2 more answers

1) 0.52 + 1.6 + 8.26 =

Nadya [2.5K]

Step-by-step explanation:

1 10.38

2 16.77

3 11.499

4 11.486

5 0.0947

6 30.52875

7 525

8 32.57862069

9 16.5625

5 0

2 years ago

Cell Phone Company A charges $20 each month plus $0.03 per text. Cell Phone Company B charges $5 each month plus $0.07 per text.

Ymorist [56]

Cost of company A=monthlyfee+text=20+0.03t where t=number of texts
A=20+0.03t

company b is monthlyfee+texts=5+0.07t
B=5+0.07t

1.
c=20+0.03t
c=5+0.07t
using c for cost is confusing because we can't tell which is for which company

2.

equal, means the costs are equal so
20+0.03t=5+0.07t
minus 5 from both sides
15+0.03t=0.07t
minus 0.03t from both sides
15=0.04t
divide both sides by 0.04
375=t
the answer is 375 texts

4 0

3 years ago

Read 2 more answers

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago