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2 years ago

13

Do you think it would be difficult to convert between the English Customary System and the Metric System? For example, if I want

ed to know how many grams were in a pound, what would I have to do?

1 answer:

tester [92]2 years ago

7 0

Answer:

1 pound is 453.6 grams appox

Step-by-step explanation:

to calculate how many grams is 2 pounds, u can do 2x453.6 which would be 907.2

for all others i have an attached file :)

You might be interested in

One of the largest pumpkins ever grown weighed about 3/4 ton.How many pounds did the pumpkin weigh?

atroni [7]

1500 pounds since a ton I 2000 just multiply it by 3/4

6 0

2 years ago

Anna is making a banner out of 4 congruent triangles as shown below. How much blue trim will she need for each side

Ede4ka [16]

Answer:

The length of each blue trim is 17.2 inches

Step-by-step explanation:

Given

See attachment for banner

Required

The length of each blue trim

Since all 4 triangles are equal, then the dimension of 1 triangle is:

$Side\ 1 = 10$

$Side\ 2 = 14$

The hypotenuse (x) of the triangle blue trim is represented by the blue trim

So:

$x^2 = Side\ 1^2 + Side\ 2^2$

This gives

$x^2 = 10^2 + 14^2$

$x^2 = 100 + 196$

$x^2 = 296$

Take square root

$x = \sqrt{296$

$x = 17.2in$

3 0

3 years ago

WILL GIVE BRAINLIEST AND THANKS. find the value of x

san4es73 [151]

Answer:

180 - 36? I dont exactly remember how to do this

Step-by-step explanation:

Sorry if its not right

5 0

2 years ago

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago

The wheels on Darlene's car have on 11-inch radius. If the wheels are rotating at a rate of 378 rpm, find the linear speed in mi

elena-s [515]

Answer:

The linear speed in which Darlene is traveling is 24.74 miles per hour.

Step-by-step explanation:

The wheel experiments rolling, which is a combination of translation and rotation. The point where linear speed happens is located at geometrical center of the wheel and instantaneous center of rotation is located at the point of contact between wheel and ground. The linear speed ( $v$ ), measured in inches per second, is defined by following expression:

$v = R\cdot \omega$ (1)

Where:

$R$ - Radius of the wheel, measured in inches.

$\omega$ - Angular speed, measured in radians per second.

If we know that $R = 11\,in$ and $\omega \approx 39.584\,\frac{rad}{s}$ , then the linear speed, measured in miles per hour, in which Darlene is traveling is:

$v = 11\,in\times \frac{1\,mi}{63360\,in} \times 39.584\,\frac{rad}{s}\times \frac{3600\,s}{h}$

$v \approx 24.74\,\frac{mi}{h}$

The linear speed in which Darlene is traveling is 24.74 miles per hour.

8 0

3 years ago