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4 years ago

Points A(–3, –2) and C(4, 5) are the endpoints of a diagonal of a square. Give the coordinates of the other two vertices.

Mathematics

1 answer:

blagie [28]4 years ago

6 0

Answer:

B(4,-2) D(-3,5)

Step-by-step explanation:

you must graph the points and project the lines in X and Y until they intersect and form a square

I attached an image

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Write the number that comes next in the sequence. 8,380; 8,290; 8,200;

Inessa [10]

Answer:

8,110

Step-by-step explanation:

It's falling by 90 each time

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3 years ago

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The top of a ladder rest at a height of 15 feet against the side of a house. If the base of a ladder is 6 ft from the house what

MariettaO [177]

Answer:

16 feet

Step-by-step explanation:

We use Pythagorean Theorem to find the height.

leg of triangle= height of ladder = 15 feet

The hypotenuse of the formed triangle = length of ladder = L

base = 6 feet

What is length of ladder?

L^2 = 15^2 + 6^2 = 225 + 36 = 261

L = root(261) = 16.155 feet

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3 years ago

How many 1/2 s are in 4 ?

Mazyrski [523]

Answer:

Step-by-step explanation:

To find this multiply 4/1 by 2/1.

4/1 x 2/1 = 8/1 = 8.

I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

7 0

3 years ago

Read 2 more answers

Suppose that v1(t)=100cos(ωt)and v2(t)=100sin(ωt). Use phasors to reduce the sum vs(t)=v1(t)+v2(t) to a single term of the form

GuDViN [60]

Answer:

$vs(t)=152.08cos(\omega t -45)=152.08\angle-45$

Step-by-step explanation:

A tension expressed in form:

$V(t)=Vm*cos(\omega t+\phi)$

It can be converted directly into a fasor:

$V(t)=Vm\angle \phi$

If the function is a positive sine we subtract 90 °

$V(t)=Vm*sin(\omega t+\phi)=Vm*cos(\omega t+\phi-90)$

So:

$v1(t)=100cos(\omega t)= 100\angle0\\v2(t)=100sin(\omega t)=100cos(\omega t-\ph90)=100\angle -90$

$vs(t)=v1(t)+v2(t)=100\angle 0 +100\angle-90=152.0811931\angle-45\approx152.08\angle-45$

Therefore:

$vs(t)=152.08\angle-45=152.08cos(\omega t -45)$

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4 years ago

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Ierofanga [76]

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3 years ago