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masha68 [24]
3 years ago
7

Calculate the area of the shape​

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
6 0

Answer:

9m

Step-by-step explanation:

The formula to find the area of a square or rectangle is to multiply the length by the width. So, we multiply 2m by 4.5m. That equals 9m. I hope this helps! Please mark me brainliest!

amm18123 years ago
5 0

Answer:

9m

Step-by-step explanation:

2m×4.5m=9

4.5+4.5=9

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The equations of four lines are given. Identify which lines are perpendicular.
Maksim231197 [3]

Answer:

lines 1 and 3

Step-by-step explanation:

y = 2 is a horizontal line parallel to the x- axis

x = - 4 is a vertical line parallel to the y- axis

Then these 2 lines are perpendicular to each other

y = 15x - 3 ( in the form y = mx + c ) with m = 15

y + 1 = - 5(x + 2) ( in the form y - b = m(x - a) with m = - 5

For the lines to be perpendicular the product of their slopes = - 1

However

15 × - 5 = - 75 ≠ - 1

The 2 lines 1 and 3 are perpendicular

3 0
3 years ago
Find the experimental probability that the first three students to arrive at school are boys
julia-pushkina [17]

Answer:

that depends

Step-by-step explanation:

how many students are in the school all together, see you would do the 3 boys over the number of people (or students) in the school

hoped this helped a little

5 0
4 years ago
Solve using the quadratic equation.<br> –3x2 - 4x + 10 = 0
d1i1m1o1n [39]

Answer:

x =(-4-√136)/6=(-2-√ 34 )/3= -2.610

or

x =(-4+√136)/6=(-2+√ 34 )/3= 1.277

Step-by-step explanation:

8 0
3 years ago
Write an equation in point – slope form from the line through the given point with the given slope. (-10, -1); m=-1
Alexxandr [17]

Answer:

y = -x - 11

Step-by-step explanation:

We know the equation must be: y = -x + c

However we do not know what "C" is. If we plug in our x and y values then we can find out the value of C.

So:

-1 = -(-10) + c

-1 = 10 + c

so c = -11

So our equation is:

y = -x - 11

5 0
3 years ago
If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?
ohaa [14]

Answer:

The greatest prime factor of k is 11.

Step-by-step explanation:

∵  LCM( 2, 15) = 30,

Thus, the even multiple of 15 must be multiple of 30,

That is, the even multiple of 15 between 295 and 615 are

300, 330, 360,..........600

Which is an AP,

Having first term, a = 300,

Common difference, d = 30,

If n be the number terms,

Last term = a+(n-1)d

=300+(n-1)30

=270 + 30n

\implies 270+30n = 600\implies 30n = 330\implies n = 11

Hence, the sum of the all even multiple of 15 from 295 to 615,

S_{11}=\frac{11}{2}(2(300)+(11-1)30)=\frac{11}{2}(600+300)=\frac{11}{2}(900)=11\times 450

According to the question,

S_{11}=k

⇒ k = 11 × 450 = 11 × 5 × 5 × 3 × 3 × 2

Hence, the greatest prime factor of k is 11.

3 0
3 years ago
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