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melisa1 [442]
4 years ago
10

Please do this if your smart please dont guess this is an important anwser​

Mathematics
1 answer:
Nataliya [291]4 years ago
4 0

Answer:

J''(- 3, 3 )

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x )

Thus J(1, 1 ) → J'(- 1, 1 )

Under a dilatation with scale factor 3, multiply each of the coordinates of J' by 3, that is

J'' = (3(- 1), 3(1)) = (- 3, 3 )

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You decide to buy a 60 in plasma-screen tv. you borrow $4300 from the electronics store at 1.8% per month simple interest, with
const2013 [10]

4300 is your principal

1.8% interest rate (change it into decimal form; 0.018)

2 years is our time

We'll now multiply our variables.

4300 x 0.018 x 2 = 154.80

This shows that in 2 years at their interest rate that they have earned 154.80.

We will add the $154.80 to the original borrowed amount of 4300.

$4454.80 will be how much they will owe after 2 years.


To find out the amount in 15 months we'll divide 15 months by 12 and we get 1.25 and now we multiply the numbers again.

4300 x 0.018 x 1.25 = 96.75

They will pay $96.75 in interest within 15 months and we'll add it to the 4300.

$4397 will be how much they will pay after 15 months.


7 0
4 years ago
What is the product written in scientific notation? (1.18×10^3)⋅(9.1×10^−6)
vladimir2022 [97]

Answer:

1.0738 x 10^-2

Step-by-step explanation:

5 0
4 years ago
Prove that if the set of vectors {u1, u2, u3} is linearly dependent and u4 is any vector, then the set {u1, u2, u3, u4} is linea
Lesechka [4]

Answer with Step-by-step explanation:

We are given that the set of vectors{u_1,u_2,u_3} is lineraly dependent set .

We have to prove that the set {u_1,u_2,u_3,u_4} is linearly dependent .

Linearly dependent vectors : If the vectors u_1,u_2.u_3,u_4

are linearly  dependent therefore the linear combination

a_1u_1+a_2u_2+a_3u_3+a_4u_4=0

Then ,there exit a scalar which is not equal to zero .

Let a_1\neq0 then the vector u_1 will be zero and remaining  other vectors are not zero.

Proof:

When u_1,u_2,u_3 are linearly dependent vectors therefore, linear combination of vectors of given set

a_1u_1+a_2u_2+a_3u_3=0

By definition of linearly dependent vector

There exist a scalar which is not equal to zero.

Suppose a_1\neq 0 then  u_1=0

The linear combination of the set {u_1,u_2,u_3,u_4}

a_1u_1+a_2u_2+a_3u_3+a_4u_4=0

When a_1\neq0\; and\; u_1=0

Therefore,the set {u_1,u_2,u_3,u_4} is linearly dependent because it contain a vector which is zero.

Hence, proved .

5 0
3 years ago
What is a synthetic division used for
Ksivusya [100]

Answer:

pooping

Step-by-step explanation:

1.poop

2.wipe.

3.eat

4.sleep

5 0
3 years ago
$1250 + 46x greater than or equal to 7,000
lyudmila [28]

ANSWER: x=$125 to make it equal to 7,000

working the problem:

7,000-1,250=5,750

5,750÷46=125

1,250+46×125 is greater than or equal to 7,000

5 0
3 years ago
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