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amid [387]
1 year ago
14

A line is drawn through (–7, 11) and (8, –9). The equation y – 11 = y minus 11 equals StartFraction negative 4 Over 3 EndFractio

n left-parenthesis x plus 7 right-parenthesis.(x + 7) is written to represent the line. Which equations also represent the line? Check all that apply.
y = y equals StartFraction negative 4 Over 3 EndFraction left-parenthesis x plus StartFraction 5 Over 3 EndFraction.x +
3y = –4x + 40
4x + y = 21
4x + 3y = 5
–4x + 3y = 17
Mathematics
2 answers:
yaroslaw [1]1 year ago
6 0

⇒Given equation of line Passing through  (–7, 11) and (8, –9) is given by

       y-11= \frac{-4}{3}(x+7)

⇒Equation of line Passing through  (–7, 11) and (8, –9) is given by

   \rightarrow \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\\rightarrow \frac{y-11}{x-(-7)}=\frac{-9-11}{8-(-7)}\\\\\rightarrow \frac{y-11}{x+7}=\frac{-20}{15}\\\\\rightarrow \frac{y-11}{x+7}=\frac{-4}{3}\\\\\rightarrow 3y-33=-4x-28\\\\\rightarrow 4x+3y=33-28\\\\ \rightarrow 4x+3y=5

Option C

          4x+3y=5

Pachacha [2.7K]1 year ago
5 0

Answer:

A and C

Step-by-step explanation:

Guest
1 year ago
It’s Aand D
Guest
10 months ago
no its not they were right
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Triangle H N K is shown. Angle H N K is 90 degrees. The length of hypotenuse H K is n, the length of H N is 12, and the length o
Mazyrski [523]

Answer:

13

Step-by-step explanation:

From the question, we are given a triangle HNK with an angle of 90°

The length of hypotenuse H K is n,

the length of HN is 12

the length of N K is 6.

From the above values, obtained in the question, we can see that this is a right angled triangle.

We are asked to find the length of the hypotenuse.

We can use Pythagoras Theorem of solve for this.

c² = a² + b²

where c = HK = n

a = NK = 6

b = HN = 12

c² = 6² + 12²

c² = 36 + 144

c² = 180

c = √180

c = 13.416407865

Approximately to the nearest whole number = 13

Therefore the value of HK = n = 13

We can also use Law of Cosines as given in the question to solve for this.

a² = b² + c² - 2ac × Cos A

where c = HK = n

a = NK = 6

b = HN = 12

Hence

c² = a² + b² - 2ab × Cos C

c = √ (a² + b² - 2ab × Cos C)

Where C = 90

c = √ 6² + 12² - 2 × 6 × 12 × Cos 90

c = 13.42

Approximately to the nearest whole number ≈ 13

Therefore the value of HK = n = 13

8 0
1 year ago
T varies inversely as the square root of u.<br>t = 3 when u = 4.<br>Find t when u = 49.​
RoseWind [281]

Answer:

t = \frac{6}{7}

Step-by-step explanation:

Given that t varies inversely as the square root of u then the equation relating them is

t = \frac{k}{\sqrt{u} } ← k is the constant of variation

To find k use the condition t = 3 when u = 4, thus

k = t\sqrt{u} = 3 × \sqrt{4} = 3 × 2 = 6

t = \frac{6}{\sqrt{u} } ← equation of variation

When u = 49, then

t = \frac{6}{\sqrt{49} } = \frac{6}{7}

6 0
1 year ago
Which polynomial correctly combines the like terms and expresses the given polynomial in standard form?
STatiana [176]

Answer:

D. 3x^4 + x^3y - 8x^2y^2 + 9xy^3 - 13y^4

Step-by-step explanation:

9xy^3 - 4y^4 - 10x^2y^2 + x^3y + 3x^4 + 2x^2y^2 - 9y^4 =

= 9xy^3 - 13y^4 - 8x^2y^2 + x^3y + 3x^4

= 3x^4 + x^3y - 8x^2y^2 + 9xy^3 - 13y^4

4 0
1 year ago
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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each equation with the operation yo
Tanzania [10]

The quadratic equations and their solutions are;

9 ± √33 /4 = 2x² - 9x + 6.

4 ± √6 /2 = 2x² - 8x + 5.

9 ± √89 /4 = 2x² - 9x - 1.

4 ± √22 /2 = 2x² - 8x - 3.

Explanation:

Any quadratic equation of the form, ax² + bx + c = 0 can be solved using the formula x = -b ± √b² - 4ac / 2a. Here a, b, and c are the coefficients of the x², x, and the numeric term respectively.

We have to solve all of the five equations to be able to match the equations with their solutions.

2x² - 8x + 5, here a = 2, b = -8, c = 5.                                                  x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(5) / 2(2) = 8 ± √64 - 40/4.     24 can also be written as 4 × 6 and √4 = 2. So                                                                                     x = 8 ± 2√6 / 2×2= 4±√6/2.

2x² - 10x + 3, here a = 2, b = -10, c = 3.                                                   x =-b ± √b² - 4ac / 2a =-(-10) ± √(-10)² - 4(2)(3) / 2(4) = 10 ± √100 + 24/4. 124 can also be written as 4 × 31 and √4 = 2. So                                                                              x = 10 ± 2√31 / 2×2 = 5 ± √31 /2.

2x² - 8x - 3, here a = 2, b = -8, c = -3.                                                    x = -b ± √b² - 4ac / 2a = -(-8) ± √(-8)² - 4(2)(-3) / 2(2) = 8 ± √64 + 24/4.     88 can also be written as 4 × 22 and √4 = 2. So                                                                             x = 8 ± 2√22 / 2×2 = 4± √22/2.

2x² - 9x - 1, here a = 2, b = -9, c = -1.                                                     x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(-1) / 2(2) = 9 ± √81 + 8/4.                                          x = 9 ± √89 / 4.

2x² - 9x + 6, here a = 2, b = -9, c = 6.                                                    x = -b ± √b² - 4ac / 2a = -(-9) ± √(-9)² - 4(2)(6) / 2(2) = 9 ± √81 - 48/4.                                                                             x = 9 ± √33 / 4

To match we solve the monomials.

1. -15u^3 + 5u^3

Adding

-15u^3 + 5u^3=-10u^3

2.  10u^3 +(-5u^3)

Adding

10u^3-5u^3=5u^3

3. 10u^3 + 5u^3

Adding

10u^3 + 5u^3=15u^3

4.  5u^3+ (-10u^3)

Adding

5u^3-10u^3 =-5u^3

Two separate ways to find the answers.

7 0
1 year ago
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