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3 years ago

If point (a, b) is in the second quadrant, in what quadrant is (a,-b). (-a, b), (-a,-b), (b, a)

Mathematics

1 answer:

MakcuM [25]3 years ago

5 0

Answer:

Step-by-step explanation:

by using these equations we can find the location by rotating and counting the quadrants. if (a,b) is the og pair, then

(a,b)>(a,-b)= quadrant 3

(a,b)>(-a,b)=quadrant 1

(a.b)>(-a,-b)= quadrant 4

(a,b)>(b,a)= quadrant 2

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Answer:

Step-by-step explanation:

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3 years ago

Mrs. Swanson gives out only one type of candy for Halloween. The local

xxTIMURxx [149]

Answer:

Cost of 18 pounds candy = $22.5

Cost of 10 pounds candy = $12.5

Cost of 1 pound of candies = $1.25

Step-by-step explanation:

6 pounds of butterscotch candies cost $7.50

⇔ 1 pound of same candy costs $\frac{7.50}{6} =$ $1.25

Now, we need to find the cost of 18 pounds of candy:

as, the cost of 6 pounds candy = $7.50

⇔ Cost of 6 x 3 = 18 pounds candy = $7.50 x 3 = $22.5

Cost of 10 pounds candy = 10 x (Cost of 1 pound candy)

= 10 x ($1.25) = $12.5

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= $1.25

8 0

3 years ago

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mart [117]

Here is your answer. I hope you like it

5 0

3 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

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4 years ago

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IgorLugansk [536]

1) question 18 of 20
-7x/5-(4y)=7
4y=-7x/5-(7)
y=-7x/20-(7/4).

if x=0 ⇒y=-7*0/20-7/4=-7/4 ⇒y-intercept is-7/4
fi y=0 ⇒ -7x/20-(7/4)=0
-7x/20=7/4
x=(7*20) / [4*-(7)]=-5 ⇒x-intercept is -5

Solution: D) x-intercept is -5 ; y-intercept is -7/4

Question 19 of 20
P₁=(x₁,y₁)
P₂=(x₂,y₂)
m=slope

m=(y₂-y₁) / (x₂-x₁)

Then:
A(1,7)
B(10,1)

m=(1-7) / (10-1)=-6/9=-2/3.

Solution: B)-2/3

Question 20 of 20:
y-y₀=m.(x-x₀)
A(4,3)
B(-4,-2)

m=(-2-3) / (-4-4)=-5/-8=5/8

y-3=(5/8).(x-4)
y=5x/8-(20/8)+3
y=5x/8-(20/8)+24/8
y=5x/8+(4/8)
y=5x/8+1/2

solution: B) y=5x/8+1/2

5 0

3 years ago