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2 years ago

A 1st degree polynomial function whose graph intercepts the y-axis at 8.

Mathematics

1 answer:

asambeis [7]2 years ago

4 0

Your answer can be anything in the form y = mx+8 where you replace m with any real number.

You start with y = mx+b, and then replace the b with the y intercept 8.

The y intercept is where the polynomial crosses the y axis.

The value of m does not matter. So you could have y = 2x+8 or y = 3x+8 for instance. Replace m with whatever your favorite number is.

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In △EKL, m∠K = 90º, m∠E = 25º, EK = 3 cm, KH - altitude. Find EH.

laiz [17]

Answer:

EH = 3.31

Step-by-step explanation:

We have been given a right angle triangle EKL. As KH has been given as the altitude (perpendicular) of the right angled triangle, and K is the right angle, we can say that EK is tthe base of the triangle and EH is the only side lleft, which is the hypotenuse of the triangle.

Where,

EK = Base = 3

KH = perpendicular altitude

EH = Hypotenuse

m<K = 90

m<E = 25

We know that

cosθ = Base/ Hypotenuse

cos 25 = 3/ EH

EH = 3/cos25

EH = 3.31

Perpendicular alitutude can also be calculated by using the formula for tanθ.

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2 years ago

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Wittaler [7]

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7 0

2 years ago

On a coordinate plane, a triangle has points (negative 5, 1), (2, 1), (2, negative 1).

daser333 [38]

Answer:

Step-by-step explanation:

Given a triangle has points:

(-5,1),(2,1), (2,-1)

Let us label the points:

A(2,1),

B(-5,1) and

C(2,-1)

To find:

Distance between (−5, 1) and (2, −1) i.e. BC.

Horizontal leg AB and

Vertical leg, AC.

Solution:

Please refer to the attached diagram for the labeling of the points on xy coordinate plane.

We can simply use Distance formula here, to find the distance between two coordinates.

Distance formula :

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

For BC:

$x_2 = 2\\x_1 = -5\\y_2 = -1\\y_1 = 1$

$BC = \sqrt{(2--5)^2+(-1-1)^2} = \sqrt{63}$

Horizontal leg, AC:

$x_2 = -5\\x_1 = 2\\y_2 = 1\\y_1 = 1$

$AC = \sqrt{(2-(-5))^2+(1-1)^2} = 7$

Vertical Leg, AB:

$x_2 = 2\\x_1 = 2\\y_2 = -1\\y_1 = 1$

$AB = \sqrt{(2-2)^2+(-1-1)^2} = 2$

5 0

2 years ago

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Juan put three square tiles with sides 8 centimeters, 10 centimeters, and x centimeters together so that they form a right trian

Elan Coil [88]

Since the area of a square is equal to the square of one of its side's length, then the area should be equivalent to $x^{2}$ .
$A = x^{2}$ ---> equation (1)
By using pythagoras rule which states that the $x^{2} = hyp^2 - opposite^2$ ---> equation (2)
where the opposite side's length is 8 and the hypotenuse side's length is 10
by substituting by the values in equation (2) therefore,
$x^{2} = 10^{2} - 8^{2}$ substitute this value in equation (1) then
$A = x^{2} = 10^{2} -8^{2}$
where A is the area of the square whose side is x

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2 years ago

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Jed entered the following values into the TVM Solver on his graphing calculator.

Elden [556K]

Answer: 17.7

Step-by-step explanation:

69/ 3.9%•100= 17.69= 17.7

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2 years ago

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