What is 30 3/4 × 9 1/3 ?

Please help me i need help

kenny6666 [7]

Answer:

H

E

A

R

T

S

Step-by-step explanation:

easy

8 0

3 years ago

Just Need A Few Questions Answered To Finish My Quiz, Any Help Would Be Much Appreciated.

kvv77 [185]

Given the triangle

PQR

with points

P(8,0)

Q(6,2)

R(-2,-4)

And the triangle

P'Q'R'

with points

P'(4,0)

Q'(3,1)

R'(-1,-2)

Part A. Scale factor

Using the vertex

P( 8, 0)

P'(4,0)

the dilatation factor is given by

$\frac{4}{8}=\frac{1}{2}$

The triangle has a dilatation factor of 1/2

Part B:

P''Q''R'' after using P'Q'R' reflected about the y axis

to make a reflection over the y axis

coordinates (x,y) turn into coordinates (-x,y)

as follows

$P^{\prime}(4,0)\rightarrow P^{\prime\prime}(-4,0)$ $Q^{\prime}(3,1)\rightarrow Q^{\prime\prime}(-3,1)$ $R^{\prime}(-1,-2)\rightarrow R^{\prime\prime}(1,-2)$

Then triangle P''Q''R'' has coordinates

P''(-4,0)

Q''(-3,1)

R''(1,-2)

Part C:

PQR is congruent to P''Q''R''?

Congruent triangles are triangles that have the same size and shape. This means that the corresponding sides are equal and the corresponding angles are equal.

Then the triangles are not congruent

8 0

1 year ago

The formula for a trapezoid relates the area A, the two bases, a and b, and the height, h. A=(a+b)/2 h Solve for a.

yKpoI14uk [10]

Answer:

$a=\frac{2A}{h}-b$

Step-by-step explanation:

Hello, I think I can help you with this.

According to the information provided in the problem, the area of a trapezoid is given by

$A=\frac{a+b}{2} *h\\$

Step 1

solve for a, it means isolate a

$A=\frac{a+b}{2} *h\\divide\ both\ sides\ by\ h\\\\\frac{A}{h}= \frac{\frac{a+b}{2} *h}{h} \\\frac{A}{h}=\frac{a+b}{2}\\\\Now, multiply\ both\ sides\ by\ two\\\\ 2*\frac{A}{h}=2*\frac{a+b}{2}\\2*\frac{A}{h}=a+b\\\frac{2A}{h}=a+b\\\\subtract\ b\ for\ both\ sides\\\frac{2A}{h}-b=a+b-b\\a=\frac{2A}{h}-b$

Have a good day.

5 0

3 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

3 years ago

Is the point (1,5) a solution to the equation below?

allochka39001 [22]

Answer:

A. No

Step-by-step explanation:

y = 2x + 4

(5) = 2(1) + 4

5 = 2 + 4

5 = 6 (false statement)

7 0

3 years ago