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4 years ago

I need help with this

Mathematics

2 answers:

jasenka [17]4 years ago

8 0

Trey = .4

Luis = .7

Ethan = .5

Roderick = .6

Hope this helps!

Schach [20]4 years ago

6 0

The answer is B Luis because he only read 1.4 pages a minute.

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Rod works as a sales clerk at a department store. He is paid a weekly salary of $300 and a commission of 3% on all sales over $5

exis [7]

7 777777777777777777777777777777777

5 0

4 years ago

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Find a₁0 given that a7 = 6 and d = 3 in the arithmetic sequence.

levacccp [35]

The tenth term of the arithmetic sequence will be 15. Then the correct option is A.

<h3>What is the arithmetic sequence?</h3>

Let a₁ be the first term and d is the common difference between the terms. Then the nth term will be given as

$\rm a_n = a_1 + (n - 1)d$

We have

a₇ = 6

d = 3

n = 7

Then the first term will be

a₇ = a₁ + (7 – 1)3

6 = a₁ + 18

a₁ = 6 – 18

a₁ = -12

Then the 10th term will be

a₁ = -12

d = 3

n =10

Then we have

a₁₀ = -12 + (10 – 1)3

a₁₀ = -12 + 27

a₁₀ = 15

Then the tenth term of the arithmetic sequence will be 15.

Then the correct option is A.

More about the arithmetic sequence link is given below.

brainly.com/question/15412619

#SPJ1

7 0

2 years ago

-2 + 4 (5x + 2) pls help

harkovskaia [24]

Answer:

6 + 20x

Step-by-step explanation:

-2 + 4 (5x + 2)

-2 + 20x + 8

6 + 20x

8 0

3 years ago

Round 56.14 to the nearest tenth

kodGreya [7K]

Answer:

Step-by-step explanation:

56.1

3 0

3 years ago

Give me a complete and managed answer.

nikdorinn [45]

Answer:

Solution given:

The given equation of a line is

ax²+2hxy+by²=0

Let y=mx be any one line of

ax²+2hxy+by²=0

Let the perpendicular line of

y=mx is

x+my=0

According to the question the line x+my=0 is one line of Ax²+2Hxy +By²=0

Substituting value of y

Ax²+2Hx $\frac{ - x}{m}$ + $B \frac{x²}{m²}$ =0

Ax²- $\frac{ 2H}{m}x²$ + $B \frac{x²}{m²}$ =0

Taking LCM

we get

Ax²m²-2mHx²+Bx²=0

x²[Am²-2Hm+B]=0..............[1]

Again.

ax²+2hx(mx)+B(mx)²=0

ax²+2hmx²+Bm²x²=0

x²[a+2hm+Bm²]=0

bm²+2hn+a=0.....................[2]

Taking coefficient of equation 1 &2.

equation 1. b 2h a b 2h

equation 2.A. -2h B A -2H

Doing criss cross multiplication

ignore first coefficient and repeat first and second

again

lines are perpendicular so

$\frac{m²}{2hB}$ = $\frac{m}{aA-bB}$ = $\frac{1}{-2Hb-2hA}$

Taking 1st & 2nd ratio,we get,

m= $\frac{2hB+2Ha}{aA-bB}$ ....[*]

Taking 3rd & 2nd ratio,we get,

m= $\frac{aA-bB}{-2Hb-2hA}$ ....[#]

Again

Equating equation * &# we get;

$\frac{2hB+2Ha}{aA-bB}$ = $\frac{aA-bB}{-2Hb-2hA}$

(aA-bB)²=-4(hB+Ha)(Hb+HA)

(aA-bB)²+4(hB+Ha)(Hb+HA)=0 is a required condition.

6 0

3 years ago

Read 2 more answers