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3 years ago

Which of the following is the equation of the line with a slope 3 and y-intercept of 4?

Mathematics

2 answers:

My name is Ann [436]3 years ago

8 0

The answer is b I am pretty sure

spin [16.1K]3 years ago

4 0

Answer:

Step-by-step explanation:

-y = -3x + 4

y = 3x - 4

slope 3 y-intercept 4

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Y= 16t (squared) +120 what is t

Zina [86]

Answer:

Step-by-step explanation:

hello :

16t²+120 = y

16t² = y -120 so : t² = (y-120)/16

if : y-120 ≥ 0 t = ±√((y-120)/16)

5 0

4 years ago

TRIGONOMETRY , can you solve this plz:( even if you know one of them,plz help:)

daser333 [38]

Answer:

See below

Step-by-step explanation:

$\frac{1}{1-cos A} - \frac{1}{1+cos A} =2 cosec A cot A \\ \\ L. H. S.= \frac{1}{1-cos A} - \frac{1}{1+cos A} \\ \\ = \frac{1+cos A - (1 - cos A)}{(1-cos A)(1+cos A)}\\ \\ = \frac{1+cos A - 1 + cos A}{1-cos^{2} A}\\ \\ = \frac{2cos A }{sin^{2} A}\\ \\ =2 .\frac{1}{sin A}.\frac{cos A }{sin A}.\\ \\ = 2.cosec \: A.cot \: A \\ \\ = R.H.S$

5 0

3 years ago

Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera

Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

= $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

= $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

= $\frac{1}{2}(\sqrt{169-144})(12)$

= $\frac{1}{2}(5)(12)$

= 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

= $\frac{(1.73)(169)}{4}$

= 73.0925

≈ 73.09 cm²

Area of rectangle III = Length × width

= CF × CD

= 7 × 5

= 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2 $\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

= 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

= 237.09 cm²

7 0

3 years ago

Will Mark brainliest if answer both correctly / What is X= ? / conditional statement

harkovskaia [24]

The answer is for 1st question is “10.4” and for the second question “non of these choices are correct”.

Explanation:

For the 1st question:

tan = opposite / adjacent

tan 30 = x / 18

18 tan 30 = x

x = 10.4

For the 2nd question,
If a triangle as an angle of 90 degree, it is 100% a right triangle.

Hope it helps :), mark me brainliest please!

6 0

3 years ago

A cylinder has radius r and height h. A. How many times greater is the surface area of a cylinder when both dimensions are multi

Ierofanga [76]

Answer: A. Factor 2 => 4x greater

Factor 3 => 9x greater

Factor 5 => 25x greater

Step-by-step explanation: A. A cylinder is formed by 2 circles and a rectangle in the middle. That's why surface area is given by circumference of a circle, which is the length of the rectangle times height of the rectangle, i.e.:

A = 2.π.r.h

A cylinder of radius r and height h has area:

$A_{1}$ = 2πrh

If multiply both dimensions by a factor of 2:

$A_{2}$ = 2.π.2r.2h

$A_{2}$ = 8πrh

Comparing $A_{1}$ to $A_{2}$ :

$\frac{A_{2}}{A_{1}}$ = $\frac{8.\pi.rh}{2.\pi.rh}$ = 4

Doubling radius and height creates a surface area of a cylinder 4 times greater.

By factor 3:

$A_{3} = 2.\pi.3r.3h$

$A_{3} = 18.\pi.r.h$

Comparing areas:

$\frac{A_{3}}{A_{1}}$ = $\frac{18.\pi.r.h}{2.\pi.r.h}$ = 9

Multiplying by 3, gives an area 9 times bigger.

By factor 5:

$A_{5} = 2.\pi.5r.5h$

$A_{5} = 50.\pi.r.h$

Comparing:

$\frac{A_{5}}{A_{1}}$ = $\frac{50.\pi.r.h}{2.\pi.r.h}$ = 25

The new area is 25 times greater.

B. By analysing how many times greater and the factor that the dimensions are multiplied, you can notice the increase in area is factor². For example, when multiplied by a factor of 2, the new area is 4 times greater.

3 0

3 years ago