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3 years ago

15

Help pls i need it "can u find the area

2 answers:

Dmitriy789 [7]3 years ago

8 0

The answer would be 84 because 10*10=100 and .5 times 4*4 equals 16 and they are both right triangles so 100-16= 84

Orlov [11]3 years ago

5 0

ANSWER:

EXPLANATION:

. :>

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-3(1-2x)-3=-20-8x what is the work and answer

larisa86 [58]

You have to keep order of operations in mind.

1. Take care of the parentheses; distribute.
-3+6x-3=-20-8x
Simplified: 6x-6=-20-8x

2. Isolate the x variable.
Add 8x to both sides and add 6 to both sides.
14x=-14

3. Solve for x.
Divide by 14 on both sides.
x=-1

4 0

3 years ago

Please help 10 points

sergij07 [2.7K]

14 Feeders is the correct answer

7 0

3 years ago

Read 2 more answers

What Is The Circumference Of This Circle? Use 3.14 For Pi. There Is A Line Down The Middle Saying "12 cm"

larisa [96]

12 is the diameter. We need the circumference which you get by multiplying 12x2 sides which = 24 which is the circumference

4 0

3 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago

For the first 40 hours of work each week, Mr. Ericson earns

labwork [276]

C. because it’s middle and the teacher prob put the answer as c

5 0

2 years ago