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Jobisdone [24]
3 years ago
13

Can u help pls thank uu so much ?

Mathematics
1 answer:
murzikaleks [220]3 years ago
5 0
Is the question what is H?
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Russell collected 30 stones at his grand parents' house. He decided to give his little sister 1 / 5 of the stones. How many ston
STALIN [3.7K]
He gives his sister 6 stones

5 0
3 years ago
What is the volume of the right prism?
Bumek [7]
Volume of a right prism = Area of the base * height

the base is the right triangle.
Area of a right  triangle = ab/2
a = 15 inches ; b = 8 inches

Area of a right triangle = (15in * 8in)/2 = 120 in² / 2 = 60 in²

V = B * h
V = 60 in² * 10 in 
V = 600 in³  Choice B.
7 0
4 years ago
One integer is 8​ times another. if the product of the two integers is 288, then find the integers.
Anarel [89]
One integer is x
another integer is y

x=8y
xy=288

(8y)(y)=288
8y^2=288
Divide both sides by 8
y^2= 36
The square root of 36 is 6
y=6

x=8y = 8(6) = 48
x=48
7 0
3 years ago
The expression 6+16k factored using the GCF is what???????
tatiyna

Answer:

2(3 + 8k)

Step-by-step explanation:

given : 6+16k

note that the GCF of 6 and 16 is 2, hence we can factor 2 out of the expression

6+16k

= (2)(3) + (2)(8k)

= 2(3 + 8k)

4 0
3 years ago
Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

2) Assume that f(k) is divisible by 5, f(k)=(k)^{5}-k

3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)      (divisible by 5)

add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

3 0
3 years ago
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