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2 years ago

I need help ASAP please help! 10th grade algebra

Mathematics

1 answer:

Sholpan [36]2 years ago

5 0

This is 10th grade I’m in regular 8th and can do this w no problem

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The image represents a rectangular patio that has a length equal to 4 feet more than the width. What is the perimeter of the pat

IrinaK [193]

Answer:

4w+8 feet

Step-by-step explanation:

The perimeter of a rectangle is the sum of the lengths of its sides. Since opposite sides are identical in length, it can be computed as double the sum of adjacent side lengths. Here, the lengths are in feet.

P = 2(L+W)

P = 2((w+4) +w)

P = 4w +8

The perimeter of the patio is 4w+8 feet.

4 0

2 years ago

What is the slope of the equation y – 3 = –4(x – 5)?

Gemiola [76]

Find the x and y intercepts or any other 2 points. (I advise you finding the y intercept though.)

then calculate for slope using:
y(2) - y(1)
________
x(2) - x(1)

The (2) and (1) are not exponents they are small numbers at the bottom of the numbers.

The slope is: 4

8 0

3 years ago

1. What is the formula for Pythagorean Theorem

Debora [2.8K]

Answer:

The formula is a^2 + b^2 = c^2

Step-by-step explanation:

A and B are legs and C is the hypothunisis.

The theorem is used for triangles.

4^2 + 5^2 = c^2

16 + 25 = c^2

41 = c^2

sqr(41) = sqr(c)^2

c = 6.403124237

4 0

2 years ago

How do you find the ratio of 40:28

ratelena [41]

First reduce it.
10:7 In a sense that is about as far down as you can go
You could however make it 1 3/7 to 1

6 0

3 years ago

2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.

likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

$y = ^+_- \sqrt{-x^2+4x+4} -5$

Thus

$f(x) = \sqrt{-x^2+4x+4} - 5$

$g(x) = -\sqrt{-x^2+4x+4} - 5$

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

$\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0$

The quadratic has roots

$\frac{52 ^+_-\sqrt{1664}}{26}$

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0

2 years ago