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3 years ago

Write the equation of the line that passes through the given points. (0.5) and (3.9)

Mathematics

1 answer:

Romashka [77]3 years ago

4 0

Answer:

y=4/3x+5

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(9-5)/(3-0)

m=4/3

y-y1=m(x-x1)

y-5=4/3(x-0)

y=4/3(x)+5

y=4/3x+5

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Pls help my brain hurts

Naya [18.7K]

Answer:

A) 58.5

Step-by-step explanation:

Pythagorean theorem: A squared + B squared = C squared

32 squared + 49 squared = 3425

Find the square root of 3425.

√ 3425 = 58.5234

Round to nearest tenth

58.5

6 0

3 years ago

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Answer this pleaseee !!

photoshop1234 [79]

Answer:

Step-by-step explanation:

m<F = m<B - Corresponding Angles Theorem

m<B + m<A = 180 - Adjacent Angles Theorem

180 - 105 = m<A

75 = m<A; therefore, m<A = 75.

6 0

3 years ago

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Trapezoid MNPQ is similar to Trapezoid RSTU. What is the length of the "x" on the misside side NP ? Also, what is the length of

ivolga24 [154]

Answer:

$x = 15$

$y = 25$

Step-by-step explanation:

Given

See attachment for MNPQ and RSTU

Required

Find x and y

To solve this question, we make use of equivalent ratios of corresponding side lengths.

The ratio of corresponding sides are:

$MN : RS$

$NP : ST$

$PQ : TU$

$MQ : RU$

From the attachment, we have:

$MN : RS \to 18 : 30$

$NP : ST \to x : 25$

$PQ : TU \to 15 : y$

To solve for x, we equate $MN : RS$ and $NP : ST$

$18 : 30 = x : 25$

Express as fraction

$\frac{18 }{ 30 }= \frac{x }{ 25}$

Make x the subject

$x = 25 * \frac{18 }{ 30 }$

$x = \frac{25 * 18 }{ 30 }$

$x = \frac{450}{ 30 }$

$x = 15$

To solve for y, we equate $MN : RS$ and $PQ : TU$

$18 : 30 = 15 : y$

Express as fraction

$\frac{18 }{ 30 }= \frac{15 }{ y}$

Make y the subject

$y = 15 * \frac{30 }{ 18 }$

$y = \frac{15 *30}{ 18 }$

$y = \frac{450}{ 18 }$

$y = 25$

8 0

3 years ago

During a sale, every item in a store is 80% of its regular price. The regular prices of five items are shown here find the sale

qwelly [4]

$55 this is shown by doing 80/100=0.80. and 0.80*55=this is the answer

6 0

3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

4 years ago

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