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3 years ago

Please help me with the questions below

Mathematics

2 answers:

iren2701 [21]3 years ago

6 0

Answer:

2) 3/4 of the girls did not have 1 piece bathing suits

3) 4/6 were at the kiddie pool

4) 1/3 was boys

sorry if this was wrong

Lelu [443]3 years ago

4 0

3/4 of the girls did not have a bathing suit. Since 1/4 of the girls did that would leave 3 out of the four left.

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The sequence an = one third(3)n − 1 is graphed below: coordinate plane showing the points 2, 1; 3, 3; and 4, 9 Find the average

Maru [420]

The given sequence is:

$a(n)= \frac{1}{3} (3)^{n-1}$

a(2)=1
a(3)=3
a(4)=9

We are to find the average rate of change between n=3 and n=4 for the given function.

Average rate of change = $\frac{a(4)-a(3)}{4-3} = \frac{9-3}{1}=6$

So the average rate of change for the given function from n = 3 to n = 4 is 6

8 0

3 years ago

Read 2 more answers

I need help with this!

timurjin [86]

Answer:

a=11

Step-by-step explanation:

Let's solve your equation step-by-step.

√a+5=4

Solve Square Root.

√a+5=4

a+5=42(Square both sides)

a+5=16

a+5−5=16−5(Subtract 5 from both sides)

a=11

Check answers. (Plug them in to make sure they work.)

a=11(Works in original equation)

8 0

3 years ago

Read 2 more answers

Please help and fast

Oksana_A [137]

Answer:

216 in squared

Step-by-step explanation:

using the formula, a cube has six sides so 6 squared is 36 times 6 equals to 216

4 0

3 years ago

What the area of the triangle (no troll answer or links)

Simora [160]

The answer is 270.56

Solution:
15.2*35.6=541.12
541.12/2=270.56

4 0

3 years ago

Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f

nata0808 [166]

Via Lagrange multipliers:

$L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)$
$L_x=2xy+1+\lambda y=0$
$L_y=x^2+1+\lambda x=0$
$L_\lambda=xy-5=0$

$\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y$
$xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x$

$\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}$
$xy=5\implies y=\pm\sqrt{30}$

At these points, we get local minima of $f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}$ .

- - -

Another way to do this is to make $f(x,y)$ a function independent of $y$ , which is made possible by the constraint.

$xy=5\implies y=\dfrac5x$
$\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x$
$\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}$

and so on.

8 0

3 years ago