Can you guys pls help me. I'll give brainliest

How many different ways can the line above be named? What are those names?

patriot [66]

The line above can be named in 2 different ways

7 0

3 years ago

If the original quantity is 10 and the new quantity is 13 what is the percent increase

Harman [31]

Answer:

As per the given statement:

Original quantity = 10

New quantity = 13

Change in quantity = New quantity - original quantity = 13-10 = 3

We have to find the percent increase.

$\text{Percent increase} = \frac{\text{Change in quantity}}{\text{Original quanity}} \times 100$

Substitute the given values we have;

$\text{Percent increase} = \frac{3}{10} \times 100 = 30\%$

Therefore, the percent increase is, 30%.

7 0

4 years ago

Find the angle between the following pairs of lines x^2+6xy +9y^2-4x +12y-5 =0

wlad13 [49]

Answer:

x

2

+6xy+9y

2

+4x+12y−5=0

Step-by-step explanation:

x

2

+6xy+9y

2

+4x+12y−5=0

Comparing the equation with the general equation of second degree gives

a=1,b=9,h=3,g=2,f=6,c=−5

Angle between a pair of straight lines that is tanθ=

∣

a+b

2

h

2

−ab

∣

tanθ=

∣

1+9

2

9−1×9

∣

=

9

0

tanθ=0

⇒θ=tan

−

(0)=0

∘

Angle between the pair of straight lines is zero therefore the lines are parallel.

Hence proved

3 0

3 years ago

(a) Draw the vectors a=[3, 2], b=[2,-1], and c=[7, 1]. (b) Show, by means of a sketch, that there are scalars s and t such that

UNO [17]

Answer:

4

Step-by-step explanation:

s is 2,1 t is 1

4 0

3 years ago

Determine whether each equation is True or False. In case you find a "False" equation, explain why is False.

elixir [45]

Answer:

(1) TRUE.

(2) FALSE.

(3) FALSE.

(4) TRUE.

(5) FALSE.

Step-by-step explanation:

(1) $\sqrt{32} = 2^{\frac{5}{2} }$

$2^{\frac{5}{2} } = (\sqrt{2} )^5 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 4\sqrt{2}\\\\\sqrt{32} = \sqrt{16 \ \times \ 2}\ = \ \sqrt{16} \ \times \ \sqrt{2} \ = \ 4\sqrt{2}$

Thus, the equation is TRUE.

(2) $16^{\frac{3}{8} } = 8^2$

$16^{\frac{3}{8} } =(2^4)^{\frac{3}{8} } = 2^\frac{3}{2} }= (\sqrt{2} )^3 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 2\sqrt{2} \\\\8^2 = 64$

Thus, the equation is FALSE.

(3) $4^{\frac{1}{2} } = \sqrt[4]{64}$

$4^{\frac{1}{2} }= \sqrt{4} = 2\\\\\sqrt[4]{64} = (64)^{\frac{1}{4} } = (2^6)^{\frac{1}{4} }= 2^{\frac{6}{4} } = 2^{\frac{3}{2} }=(\sqrt{2} )^3 = (\sqrt{2} \times \sqrt{2} \times \sqrt{2} ) = 2\sqrt{2}$

Thus, the equation is FALSE.

(4) $2^8 = (\sqrt[3]{16} )^6$

$2^8 = 256\\\\ (\sqrt[3]{16} )^6 = (16)^{\frac{6}{3} } = (2^4)^{\frac{6}{3} } = (2)^{\frac{24}{3} } = 2^8 = 256$

Thus, the equation is TRUE.

(5) $(\sqrt{64} )^{\frac{1}{3} } = 8^{\frac{1}{6} }\\\\$

$8^{\frac{1}{6} } = (2^3)^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} } = \sqrt{2} \\\\(\sqrt{64} )^{\frac{1}{3} } = (2^6)^{\frac{1}{3} } = 2^{\frac{6}{3} } = 2^2 = 4$

Thus, the equation is FALSE.

4 0

3 years ago