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3 years ago

What is the y-intercept of this line?

Mathematics

1 answer:

snow_lady [41]3 years ago

4 0

The y-intercept of the line is +5

Y= mx + b

The b is a variable that indicates where the line will pass through on the y-axis

You might be interested in

2 ^( 3 - 9) - 11 how do I solve this

Setler [38]

2^(3-9)-11 =

1 step: subtract the 3 and 9 to get 6

(3-9)=6

2 step: multiply 2 to its self 6 times and
you should get 64

2*2=4*2=8*2=16*2=32*2=64

3 step: subtract 11 by 64 to get 53

64-11=53

hope I help if so hit the brainliest thanks

5 0

3 years ago

You are standing 40 meters from the base of a tree that leans 8o away from you. The angle of elevation from you to the top of th

Y_Kistochka [10]

Answer:

the height of the tree is 15.49 m

Step-by-step explanation:

Step 1:

From the figure, we can determine ∠ATB by using the fact that the sum of all the angles in a triangle add up to 180°:

∠ ATB = 180° - 98° - 20°

∠ ATB = 62°

Step 2:

Therefore, using the law of sines, we can determine the height of the tree.

TB / sin(20°) = 40 / sin(62°)

TB = 40 × (sin(20°) / sin(62°))

TB = 15.49 m

Therefore, the height of the tree is 15.49 m

3 0

3 years ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

3 0

3 years ago

Heights of young adult women in the United States are approximately Normal with a mean of 64 inches and standard deviation 2.7 i

sweet-ann [11.9K]

Answer:

The proportion of all young adult women in the United States are taller than 6 feet(72 inches ) is

$P( X > 72) =0.0015233$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 64 \ inches$

The standard deviation is $\sigma = 2.7 \ inches$

Generally 6 feet is equivalent to 6 * 12 = 72 inches

Generally the proportion of all young adult women in the United States are taller than 6 feet(72 inches ) is mathematically represented as

$P( X > 72) = P(\frac{X - \mu }{\sigma } > \frac{72 - 64 }{ 2.7 } )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

$P( X > 72) = P(Z > 2.963 )$

From the z-table

The area under the normal curve to the right corresponding to 2.963 is

$P(Z > 2.963 ) = 0.0015233$

=> $P( X > 72) =0.0015233$

3 0

3 years ago

Find the equation of a line parallel to y=-10x-5 passing through the coordinate (-3,5)

Allushta [10]

With a given parallel line and a given point on the line
we can use the point-line method: y-y0=m(x-x0)
where
y=mx+k is the given line, and
(x0,y0) is the given point.

Here
m=-10, k=-5, (x0,y0)=(-3,5)

=> the required line L is given by:
L: y-5=-10(x-(-3))

on simplification
L: y=-10x-30+5
L: y=-10x-25

4 0

4 years ago