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4 years ago

Tortoise a walks 52. 0 ft./h and told us B walks 12 in./m which tortoise travels faster

Mathematics

1 answer:

makvit [3.9K]4 years ago

8 0

Tortoise A.
52 ft converted to in is 624.

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Find the area of the composite figure.

vivado [14]

So in fig 1 :

L = 14m
W = 9m

Area = L × W
Area = 9 × 14
Area = 126 m²

In fig 2 :

L = 7m
W = 16 - 9 = 7m

Area= L × W
Area = 7 × 7
Area =49m²

TA = 126 + 49
TA = 175 m²

4 0

2 years ago

Plz help geometry!!!

KIM [24]

Answer:

35 love

Step-by-step explanation:

7 0

3 years ago

A fuel-tanker's tank is 50 feet long and has a diameter of seven feet. Determine the surface area of the fuel tank.

Bezzdna [24]

Answer:

1176.679ft²

Step-by-step explanation:

<u>The question is on surface area of a cylindrical fuel tank</u>

Equation for surface area of a cylinder

S.A = 2 $\pi$ rh +2 $\pi$ r²

where r is the radius of cylinder and h is height of cylinder

Diameter=7'⇒r=3.5' and h=50'

Surface Area= 2×3.142×3.5×50 + 2×3.142×3.5×3.5

S.A=1099.7+76.979=1176.679ft²

6 0

3 years ago

Read 2 more answers

Hello can somedy help me please <br> thank you in advance

alina1380 [7]

Answer:

Area of the circle=πr²

$= 3.142 \times {4}^{2} \\ = 50.272cm \\ = 50.3cm$

4 0

3 years ago

compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.

Nina [5.8K]

$\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

3 0

1 year ago