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3 years ago

13

3/7 of students at a school are boys . If there are 2282 students at the school. How many are girls .

1 answer:

Natasha2012 [34]3 years ago

7 0

If 3/7 are boys, then 4/7 have to be girls.

2282 / 7 = 326
326 x 4 = 1304

There are 1304 girls.

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Pls help being timed

Trava [24]

If you find the least common denominator, that is 143. Then, you convert 2/13=22/143 and 1/11=13/143. Next, subtract 22/143-13/143=9/143 which is B

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3 years ago

Fill in the number that makes this sentence true 6x9=(6x5)+(6x)

antiseptic1488 [7]

<span>Fill in the number that makes this sentence true 6x9=(6x5)+(6x) (6x5)</span>

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2 years ago

Joseph is creating a dilation through point B with a scale factor of 2. Which statements about the dilation are correct? Check a

swat32

Answer:

Options (1), (2) and (4).

Step-by-step explanation:

Given question is incomplete; Find the complete question in the attachment.

Triangle ABC is a right triangle.

ΔABC has been dilated by a scale factor of 2 forming ΔA'B'C'.

These are the observations,

1). A and A' will be co-linear.

2). Since the given triangle has been dilated about a point B therefore, after dilation vertex B and B' will at be same location.

3). Length of B'C' = 2BC

4). AB and BC are the segments which will lie on the sides BA' and BC'.

5). Pre-image ABC will be inside the image A'B'C'.

Therefore, Options (1), (2) and (4) will be the answer.

4 0

3 years ago

Read 2 more answers

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

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3 years ago

-4. (5 + 1) -3 + 3+ 2

antiseptic1488 [7]

Answer:

-13

Step-by-step explanation:

7 0

2 years ago