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2 years ago

Find the value of y, given that m<KLM = 135

Mathematics

2 answers:

dangina [55]2 years ago

4 0

Answer:

the value of y is, 5.5

Explanation:

GIven: $m\angle KLM=135^{\circ}$

Linear pairs is a pair of adjacent angles formed when two lines intersect.

In the figure, $m\angle kLN$ and $m\angle MLN$ forms a linear pair.

then, $m\angle KLM = m\angle KLN+m\angle MLN$ .

Substitute the values of $m\angle KLN= 47^{\circ}$ and $m\angle MLN= (16y)^{\circ}$ in above equation, to find the value of y;

⇒ $135^{\circ}=47^{\circ}+(16y)^{\circ}$ or

$16y^{\circ} =135^{\circ}-47^{\circ}$

Simplify:

$16y^{\circ} =88$

Divide 16 on both sides of an equation:

$\frac{16y}{16} =\frac{88}{16}$

Simplify:

$y=\frac{11}{2}$ or y=5.5

Therefore, the value of y = $\frac{11}{2}$ or y=5.5.

snow_tiger [21]2 years ago

3 0

I think this is right

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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

1 year ago

An arc on a circle measures 250°. Within which range is the radian measure of the central angle? 0 to StartFraction pi Over 2 En

Alex Ar [27]

Answer:

The central angle is within the range π to 3π/2

Step-by-step explanation:

To convert from degrees to radians, we multiply the angle in degrees by 180/π.

To convert from radians to degree, we multiply the angle in radians by 180°/π.

π/2 = π/2 X 180°/π= 90°

π rad = π X 180°/π= 180°

3π/2 = 3π/2 X 180°/π= 270°

2π = 2π X 180°/π= 360°

Therefore the angle 250 which is between 180 and 270 is within the range :

π to 3π/2

4 0

2 years ago

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Charles has a collection of dimes and quarters worth $1.25. He has 8 coins. Write a systems of equations to represent this situa

Dima020 [189]

Answer:

* The systems of equations are:

# d + q = 8 ⇒ (1)

# 10d + 25q = 125 ⇒ (2)

Charles has 5 dimes and 3 quarters

Step-by-step explanation:

* Lets explain how to solve the problem

- Charles has a collection of dimes and quarters worth $1.25

- He has 8 coins

* To solve the problem remember that:

# 1 dim = 10 cents

# 1 quarter = 25 cents

# 1 dollar = 100 cents

- Assume that the number of dimes is d and the number of

quarter is q

∵ Charles has 8 coins

- The number of dimes and the number of quarters equal the

number of the coins

∴ d + q = 8 ⇒ (1)

∵ 1 dime = 10 cents

∴ The value of dimes = 10 × d = 10d

∵ 1 quarter = 25 cents

∴ The value of quarters = 25 × q = 25q

∵ The collection worth $1.25

∵ 1 dollar = 100 cents

∴ The collection worth = 1.25 × 100 = 125 cents

∴ 10d + 25q = 125 ⇒ (2)

* The systems of equations are:

# d + q = 8 ⇒ (1)

# 10d + 25q = 125 ⇒ (2)

* Lets solve the equations

- Multiply equation (1) by (-10) to eliminate d

∴ -10d + -10q = -80 ⇒ (3)

- Add equations (2) and (3)

∴ 15q = 45

- Divide both sides by 15

∴ q = 3

- Substitute the value of q in equation (1) to find the value of d

∴ d + 3 = 8

- Subtract 3 from both sides

∴ d = 5

∵ d represents the number of dimes and q represents the number

of quarters

∴ Charles has 5 dimes and 3 quarters

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2 years ago

In a study in Scotland (as reported by Devlin 2009), researchers left a total of 320 wallets around Edinburgh, as though the wal

Ann [662]

Answer:

a) The observed proportion of wallets that were returned

p = 0.45625

b) 95% of confidence intervals for Population proportion

0.40168 , 0.51082)

c) The lower bound of the 95% confidence interval = 0.40168

d) The upper bound of the 95% confidence interval = 0.51082

Step-by-step explanation:

Step(i):-

Given data the researchers left a total of 320 wallets around Edinburgh, as though the wallets were lost. Each contained contact information including an address. Of the wallets, 146 were returned by the people who found them

Given sample size 'n' = 320

Given data 'x ' = 146

Sample proportion

$p = \frac{x}{n}$

$p = \frac{x}{n} = \frac{146}{320} = 0.45625$

Step(ii):-

b) 95% of confidence intervals for Population proportion

Level of significance = 95% or 0.05%

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

95% of confidence intervals for Population proportion are determined by

$(p - Z_{0.025} \frac{\sqrt{p(1-p)} }{\sqrt{n} } , p + Z_{0.025} \frac{\sqrt{p(1-p)} }{\sqrt{n} })$

$(0.45625 - 1.96\frac{\sqrt{0.45625(1-0.45625)} }{\sqrt{320} } , 0.45625 + 1.96\frac{\sqrt{0.45625(1-0.45625)} }{\sqrt{320} })$

(0.45625 - 0.05457 , 0.45625 + 0.05457)

( 0.40168 , 0.51082)

c) The lower bound of the 95% confidence interval = 0.40168

d) The upper bound of the 95% confidence interval = 0.51082

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1 year ago

What is the y-intercept of the function f(x) = –f(x) equals negative StartFraction 2 Over 9 EndFraction x plus StartFraction 1 O

lora16 [44]

Answer: Third option

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope of the line and "b" is the y-intercept.

Given the function f(x):

$f(x)=-\frac{2}{9}x+\frac{1}{3}$

Since $f(x)=y$ , you can rewrite it:

$y=-\frac{2}{9}x+\frac{1}{3}$

You can identify that:

$m=-\frac{2}{9}\\\\b=\frac{1}{3}$

Therefore, you can determine that the y-intercept of the given function is:

$b=\frac{1}{3}$

Observe that this matches with the third option.

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2 years ago

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