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2 years ago

EQUATIONS y = x2 – 1 Domain: Range: Function?

Mathematics

1 answer:

AlexFokin [52]2 years ago

8 0

Answer:

Domain: all real numbers

Range: -1,infinity

Function: yes

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Y’all know this anything about this?

Andre45 [30]

We are given:

Parallelogram CDEF
∠C = 13x - 10
∠D = 5x + 10

To Find: ∠E

Concept Used:

Adjacent angles of a Parallelogram are Supplementary

Finding the value of 'x':

Since adjacent angles of a parallelogram are Supplementary:

∠D + ∠C = 180

5x + 10 + 13x - 10 = 180 [replacing the values of angle C and D]

18x = 180

x = 10 [dividing both sides by 18]

__________________________________________________________

Finding Measure of Angle D:

We are given that:

∠D = 5x + 10

∠D = 5(10) + 10 [since x = 10]

∠D = 60°

__________________________________________________________

Finding the measure of ∠E:

∠D and ∠E are Adjacent angles

Since Adjacent angles are Supplementary:

∠D + ∠E = 180

60 + ∠E = 180

∠E = 120° [subtracting 60 from both sides]

Hence, the measure of ∠E = 120°

7 0

2 years ago

Logan spent $42 to buy 4 basketballs. What was the price per basketball

ale4655 [162]

Answer:

$10.5

Step-by-step explanation:

$42 for 4 basketballs

price of 1 basketball = 42/4

42/4 = 10.5

-Chetan K

3 0

2 years ago

Read 2 more answers

Calculate the area of a circle with a radius of 10 cm use pie

Advocard [28]

100pi or 100x3.14159

3 0

3 years ago

Jasmin has 50 marbles in a bag.20% of the marbles are blue.How many marbles are blue

Minchanka [31]

20% of 100 is 20. divide that by 2 to get 20% of 50. 20/2 is 10. 20% of 50 is 10. There are 10 blue marbles.

Hope it helped :)
Ali

6 0

3 years ago

Find the length of the curve y = 3/5x^5/3 - 3/4x^1/3 + 6 for 1 < = x < = 8. The length of the curve is . (Type an exact an

Mashutka [201]

Answer:

$\sqrt\frac{387}{20}$

Step-by-step explanation:

$Arc Length =\int\limits^a_b {\sqrt{1+(\frac{dy}{dx})^2 } } \, dx$

$y=\dfrac{3}{5}x^{\frac{5}{3}}- \dfrac{3}{4}x^{\frac{1}{3}}+6$

$\frac{dy}{dx} =x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}}$

$1+(\frac{dy}{dx})^2 }=1+(x^{\frac{2}{3}}-\dfrac{1}{4}x^{-\frac{2}{3}})^2\\=1+(x^{\frac{4}{3}}-\dfrac{1}{2}+ \dfrac{1}{16}x^{-\frac{4}{3}})$

$=\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}}$

For the Interval $1\leq x\leq 8$

Length of the Curve $=\int\limits^8_1 {\sqrt{\dfrac{1}{2}+x^{\frac{4}{3}}+ \dfrac{1}{16}x^{-\frac{4}{3}} } } \, dx\\$

Using T1-Calculator

$=\sqrt\frac{387}{20}$

3 0

3 years ago