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2 years ago

Does anyone know the answer to this¿?¿

Mathematics

1 answer:

natita [175]2 years ago

7 0

The answe is y=25 and theres 5$

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Please help with this equation problem. Find y and show your work- will give brainliest

Nikolay [14]

Answer: 8x=-2

The steps are in the photo below

3 0

2 years ago

Quadrilateral JKLM is graphed with vertices at J(-2,2), K(-1,-5), L(4,0), and M(3,7).

AveGali [126]

Answer:

Question (a)

Midpoint of a line segment:

$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Given:

J = (-2, 2)
L = (4, 0)

$\implies \textsf{midpoint of }JL=\left(\dfrac{-2+4}{2},\dfrac{2+0}{2}\right)=(1,1)$

Given:

M = (3, 7)
K = (-1, -5)

$\implies \textsf{midpoint of }MK=\left(\dfrac{3-1}{2},\dfrac{7-5}{2}\right)=(1, 1)$

Question (b)

Find slopes (gradients) of JL and MK then compare. If the product of the slopes of JL and MK equal -1, then JL and MK are perpendicular.

$\textsf{slope }m=\dfrac{y_2-y_1}{x_2-x_1}$

Given:

J = (-2, 2)
L = (4, 0)

$\implies \textsf{slope of }JL=\dfrac{0-2}{4+2}=-\dfrac13$

Given:

M = (3, 7)
K = (-1, -5)

$\implies \textsf{slope of }MK=\dfrac{-5-7}{-1-3}=3$

$\textsf{slope of }JL \times \textsf{slope of }MK=-\dfrac13 \times3=-1$

Hence segments JL and MK are perpendicular

4 0

1 year ago

A regular hexagon has sides of 2 feet. What is the area of the hexagon?

il63 [147K]

Trigonometry would help with this question.

The area of a regular hexagon is ((3√3)s^2)/2 where s is the side.

Plugging in 2 gives us 6√3 or 10.39 feet.

5 0

2 years ago

Read 2 more answers

PLEASE ANSWER! ASAP thank you!!

irina [24]

Answer:

p=75d

Step-by-step explanation:

The giraffe eats 75 pounds per day so that is the slope. So the equation is y=m(slope)x or in this case p=md

7 0

2 years ago

*****WILL GIVE BRAINLEST IF CORRECT******

PSYCHO15rus [73]

Conner's work is correct. To combine and make it simple, you Multiply:

(3^5+9)+(6^8+10) which will equal 3^14 6^18.

But Jane's work, instead of adding, Jane multiplies. So, Conner is correct.

Hope that helped!

5 0

2 years ago