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3 years ago

Write an equation of a line that is parallel to y=-6x+4 and passes through(2,10).

1 answer:

3 0

A)y=-6x+4
Slope and intercept a=-6 b=4
equation for new line:
y=cx+d
If new line is parallel c=a=-6 so
y=-6x+d
We substitude point (2,10)
10=-6*2+d /+12
d=22
so y=-6x+22
b)
y=- $\frac{2}{3}$ x-5
a=-2/3
b=-5
new line:
y=ex+f
If line is perpendicular e=- $\frac{1}{a}$
so e= $\frac{3}{2}$
we can write
y= $\frac{3}{2}x+f$
Substituting point (6,3)
3= $\frac{3}{2}*6+f$
f=3-9
f=-6
y= $\frac{3}{2}x-6$

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2 years ago

Rationalize the denominator of \frac{1 \sqrt{3}}{1-\sqrt{3}}. When you write your answer in the form A B\sqrt{C}, where A, B, an

kobusy [5.1K]

Answer:

The value of ABC is 6.

Step-by-step explanation:

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$\frac{1+\sqrt{3}}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}$

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$\frac{1^2+(\sqrt{3})^2+2\sqrt{3}}{1-3}$

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$A+B\sqrt{C}$ .... (2)

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2 years ago

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Sever21 [200]

Percentage of the area shaded = 55%

Solution:

Perimeter of the square = 40 cm

Each side of the square = $\frac{40}{4}$ = 10 cm

Area of the square = $\text{side}^2$

= 10²

= 100

Total area of the square = 100 cm²

The unshaded portion is two triangles.

Base of the smaller triangle = 10 cm – 7 cm = 3 cm

Height of the smaller triangle = 10 cm

Area of the smaller triangle = $\frac{1}{2}\times \text{base}\times \text{height}$

$$=\frac{1}{2}\times 3\times 10$

Area of the smaller triangle = 15 cm²

Base of the larger triangle = 10 cm – 4 cm = 6 cm

Height of the larger triangle = 10 cm

Area of the larger triangle = $\frac{1}{2}\times \text{base}\times \text{height}$

$$=\frac{1}{2}\times 6\times 10$

Area of the larger triangle = 30 cm²

Area of the shaded portion = Area of the square – Area of the smaller

triangle – Area of the larger triangle

= 100 cm² – 15 cm² – 30 cm²

Area of the shaded portion = 55 cm²

Percentage of the area shaded = $\frac{\text{Area of the shaded portion}}{\text{Total area of the square}}\times100\%$

$$=\frac{55}{100}\times100\%$

Percentage of the area shaded = 55%

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3 years ago