All categories

2 years ago

Which ordered pairs in the form (x, y) are solutions to the equation

Mathematics

2 answers:

Xelga [282]2 years ago

5 0

1
4
Im sure ..................

inna [77]2 years ago

4 0

Answer:

(-1, 7) and (-6, -14) are solutions

You might be interested in

Identify the domain, range, and codomain in each graph. Then use the codomain and range to determine whether the

bearhunter [10]

Answer:

Step-by-step explanation:

6 0

3 years ago

Let's say Jose deposited $2,700 into a CD account with an Annual Percentage Rate

sergey [27]

Answer:

$3283.2

Step-by-step explanation:

Given data

Principal= $2700

Rate= 4%

Time= 5 years

Required

the final Amount A

The compound interest formula is

A=P(1+r)^t

Substitute

A=2700(1+0.04)^5

A=2700(1.04)^5

A=2700*1.216

A=$3283.2

Hence the balance in the account after 5 years is $3283.2

5 0

3 years ago

The approximate height, h, in meters, traveled by golf balls hit with two different clubs over a horizontal distance of d meters

inna [77]

Answer:

a) 100 metres

b) 10 metres

Step-by-step explanation:

Given: the approximate height, h, in meters, travelled by golf balls hit with two different clubs over a horizontal distance of d meters is given by the functions: $h=-0.002d^2+0.3d\,,\,h=-0.004d^2+0.5d$

To find: distances when the ball is at the same height when either of the clubs are used. Also, to find this height

Solution:

(a)

$-0.002d^2+0.3d=-0.004d^2+0.5d\\d(-0.002d+0.3)=d(-0.004d+0.5)\\-0.002d+0.004d=0.5-0.3\\0.002d=0.2\\d=\frac{0.2}{0.002}=100\,\,metres$

(b)

Put d = 100 in $h=-0.002d^2+0.3d$

$h=-0.002(100)^2+0.3(100)\\=-20+30\\=10\,\,metres$

4 0

3 years ago

A restaraunt has 1,600 milligrams and there’s gotta be 10 people how many grams do you need

natita [175]

You need 1.6 Grams. Hope this helped :P

8 0

3 years ago

Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL

Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

=> JK = $\sqrt{(1-2)^{2} + (1-5)^{2} } = \sqrt{(-1)^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

J(2;5); L(5;2)

=> JL = $\sqrt{(5-2)^{2} + (2-5)^{2} } = \sqrt{3^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

K(1;1); L(5;2)

=> KL = $\sqrt{(5-1)^{2} + (2-1)^{2} } = \sqrt{4^{2}+1^{2} } = \sqrt{1+16}=\sqrt{17}$

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

=> QN = $\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}$

Q (-4;4); P(-7;1)

=> QP = $\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}$

N(-3;0); P(-7;1)

=> NP = $\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}$

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0

3 years ago

Read 2 more answers