Ask question

Ask question

All categories

3 years ago

12

10x is greater than or equal to 120 cm^2

1 answer:

Agata [3.3K]3 years ago

8 0

Answer:

x greater than or equal to 12cm^2

Step-by-step explanation:

You might be interested in

NEED HELP ASAP!!!!!!!!!!!!!!!!

salantis [7]

9514 1404 393

Answer:

(b) (-5, 1)

(c) (5, -7)

(d) (9, -2)

Step-by-step explanation:

The coordinate differences between the given points are ...

(4, 2) -(0, -3) = (4, 5)

The length of the line segment between the points is √(4² +5²) = √41, so this is the side of the square, not a diagonal.

The other four points that could be corners of the square are these same distances, but at right angles. To get points at right angles, the distance values can be swapped, and one of them negated.

Two of the points could be ...

(4, 2) ± (5, -4) = (9, -2) or (-1, 6)

and the other two could be ...

(0, -3) ± (5, -4) = (5, -7) or (-5, 1)

5 0

3 years ago

The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.

Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is $\dfrac{15}{2}.$

To find:

The sum of first six terms.

Solution:

We have,

$S_2=6$

$S_4=\dfrac{15}{2}$

Sum of first n terms of a GP is

$S_n=\dfrac{a(1-r^n)}{1-r}$ ...(i)

Putting n=2, we get

$S_2=\dfrac{a(1-r^2)}{1-r}$

$6=\dfrac{a(1-r)(1+r)}{1-r}$

$6=a(1+r)$ ...(ii)

Putting n=4, we get

$S_4=\dfrac{a(1-r^4)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}$

$\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}$

$\dfrac{15}{2}=6(1+r^2)$ (Using (ii))

Divide both sides by 6.

$\dfrac{15}{12}=(1+r^2)$

$\dfrac{5}{4}-1=r^2$

$\dfrac{5-4}{4}=r^2$

$\dfrac{1}{4}=r^2$

Taking square root on both sides, we get

$\pm \sqrt{\dfrac{1}{4}}=r$

$\pm \dfrac{1}{2}=r$

$\pm 0.5=r$

Case 1: If r is positive, then using (ii) we get

$6=a(1+0.5)$

$6=a(1.5)$

$\dfrac{6}{1.5}=a$

$4=a$

The sum of first 6 terms is

$S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}$

$S_6=\dfrac{4(1-0.015625)}{0.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Case 2: If r is negative, then using (ii) we get

$6=a(1-0.5)$

$6=a(0.5)$

$\dfrac{6}{0.5}=a$

$12=a$

The sum of first 6 terms is

$S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}$

$S_6=\dfrac{12(1-0.015625)}{1.5}$

$S_6=8(0.984375)$

$S_6=7.875$

Therefore, the sum of the first six terms is 7.875.

5 0

3 years ago

There are 20 machines in a factory. 7 of the machines are defective.

adelina 88 [10]

Answer:

0.1225

Step-by-step explanation:

Given

Number of Machines = 20

Defective Machines = 7

Required

Probability that two selected (with replacement) are defective.

The first step is to define an event that a machine will be defective.

Let M represent the selected machine sis defective.

P(M) = 7/20

Provided that the two selected machines are replaced;

The probability is calculated as thus

P(Both) = P(First Defect) * P(Second Defect)

From tge question, we understand that each selection is replaced before another selection is made.

This means that the probability of first selection and the probability of second selection are independent.

And as such;

P(First Defect) = P (Second Defect) = P(M) = 7/20

So;

P(Both) = P(First Defect) * P(Second Defect)

PBoth) = 7/20 * 7/20

P(Both) = 49/400

P(Both) = 0.1225

Hence, the probability that both choices will be defective machines is 0.1225

4 0

3 years ago

Please help me with this word problem.

koban [17]

I would say 6 hours but i don’t think it is, sorry this one is a real tricky one

7 0

3 years ago

Can someone help me with this question please

Ne4ueva [31]

Start with the given equation.

P1V1=P2V2

To solve for P2, move anything else on that side to the other side so that P2 is alone.

P1V1=P2V2
P2=P1V1/V2

8 0

3 years ago