All categories

2 years ago

7th grade easy math that I can’t do

Mathematics

2 answers:

statuscvo [17]2 years ago

5 0

The answer is D if u need an explanation lmk

zalisa [80]2 years ago

3 0

Answer: D

Step-by-step explanation: I’ve had this question b4 it’s D. I got it right.

You might be interested in

Round 14.6837 to the nearest HUNDREDTHS

nika2105 [10]

Answer:

it's 14.68

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Solve the inequality. -1/3w ≤ -5

omeli [17]

Answer:

$\frac{ - 1}{3} w \leqslant - 5$

$w \leqslant - 5 \div \frac{ - 1}{3}$

$w \leqslant - 5 \times - 3$

$w \leqslant 15$

3 0

2 years ago

The length of a rectangle is 5 centimeters less than twice its width. Its area is 42 square centimeters. Find the dimensions of

Mila [183]

Answer:

width=6cm, lenth is 7cm

Step-by-step explanation:

Let width of rectangle be x cm

Twice width=2x cm

length=2x-5 cm

Area of the rectangle= L× W =42cm²

(2x-5) × x =42

x(2x-5)=42

2x²-5x-42=0

solving the quadratic equation

x=6cm

width=6cm

length= (2×6)- 5 =7cm

7 0

3 years ago

What property of an even function do you see in this graph?

Brut [27]

Answer:

Step-by-step explanation:

I'm confused about what you want me to do but I know for a faction it is a function.

6 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