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3 years ago

Geometry math question no Guessing and Please show work thank you

Mathematics

1 answer:

KatRina [158]3 years ago

4 0

$m\angle DCA=m\angle BCA\ \text{therefore:}\\\\4x=6x-58\ \ \ \ |-6x\\\\-2x=-58\ \ \ |:(-2)\\\\x=29\\\\m\angle DCA=(4x)^o\to m\angle DCA=(4\cdot29)^o=116^o$

Answer: C. 116

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A video game is on sale for 25% off. The original price is $42. What is the Discount?

stealth61 [152]

$10.50 gets taken off the original price. I divided $42 by 4 because 25 is 1/4th of 100. This got me $10.50. The original price ($42) minus $10.50 equals $31.50. So the item sells for $31.50!!!!

5 0

3 years ago

The graph of y = 3x2 – 3x – 1 is shown.

likoan [24]

Step-by-step explanation:

the answer is:

0

Explaination:

3x×2-3x=6x-3x

=3x+2=2

3x=2-2

3x=0

X=0÷3=0

hope it helps

6 0

2 years ago

Arrange the equations in the correct sequence to rewrite the formula for displacement, , to find a. In the formula, d is displac

OverLord2011 [107]

Answer:

2(d-vt)=-at^2

a=2(d-vt)/t^2

at^2=2(d-vt)

Step-by-step explanation:

Arrange the equations in the correct sequence to rewrite the formula for displacement, d = vt—1/2at^2 to find a. In the formula, d is

displacement, v is final velocity, a is acceleration, and t is time.

Given the formula for calculating the displacement of a body as shown below;

d=vt - 1/2at^2

Where,

d = displacement

v = final velocity

a = acceleration

t = time

To make acceleration(a), the subject of the formula

Subtract vt from both sides of the equation

d=vt - 1/2at^2

d - vt=vt - vt - 1/2at^2

d - vt= -1/2at^2

2(d - vt) = -at^2

Divide both sides by t^2

2(d - vt) / t^2 = -at^2 / t^2

2(d - vt) / t^2 = -a

a= -2(d - vt) / t^2

a=2(vt - d) / t^2

2(vt-d)=at^2

4 0

3 years ago

PLEASE NEED THIS DONE AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!! IT MUST BE CORRECT

LuckyWell [14K]

Answer:

1)7.288 feet

2)11.6 feet

3)safe

4) 3.7 feet

5) ∅ $= tan^{-1}(1.5) = 56.31 degrees$

6)∅1 $= tan^{-1}(2) = 63.43 degrees$

∅2 $= tan^{-1}(1.2) = 50.19 degrees$

Step-by-step explanation:

1) The door barn is rectangular in shape. The length is 9 feet and The angle between diagonal and side is 39 degrees.

Applying trigonometry,

tan(39) = $\frac{opposite}{adjacent} = \frac{s}{9}$ = 0.809

Thus, s= $(9)(0.809) = 7.288 feet$

2) Applying pythagoras theorm,

$(Diagonal)^{2} = (9)^{2} + (7.288)^{2} =134.115$

Diagonal length (d) = 11.58 feet. Nearest tenth place = 11.6 feet

3) The length of ladder is 14 foot and height from ground is 13.5 feet.

Applying trigonometry,

sin(∅) = $\frac{opposite}{hypotenous} = \frac{13.5}{14}$ = 0.964

∅ = angle of elevation = $sin^{-1}(0.964)$ = 74.57 ≈ 75 degrees.

Thus tractor can climb safely.

4)Applying, pythgoras theorm,

$14^{2} = (13.5)^{2} + x^{2}$

x = $\sqrt{14^{2}-(13.5)^{2}} = 3.708$

Thus, ladder should be placed at distance 3.7 feet

5)Let angle of elevation be ∅.

tan(∅) = $\frac{30}{20} = 1.5$

∅ $= tan^{-1}(1.5) = 56.31 degrees$

6)After moving 5 feet closer to barn, Let angle of elevation for light near barn be ∅1 and for farther one be ∅2.

Thus,

tan(∅1) = $\frac{30}{15} = 2$

∅1 $= tan^{-1}(2) = 63.43 degrees$

tan(∅2) = $\frac{30}{25} = 1.2$

∅2 $= tan^{-1}(1.2) = 50.19 degrees$

3 0

3 years ago

What is 2 (log Subscript 3 Baseline 8 + log Subscript 3 Baseline z) minus log Subscript 3 Baseline (3 Superscript 4 Baseline min

Ilya [14]

Answer:

Value of expression in single logarithm is $\log_3\left(2z^2\right)$ .

Step-by-step explanation:

Given expression is,

$2\left(\log_3\left(8\right)+\log_3\left(z\right)\right)-\log_3\left(3^4-7^2\right)$

Now using logarithmic rule to solve the expression as follows,

Applying product rule of logarithmic,

$\log_c\left(a\right)+\log_c\left(b\right)=\log_c\left(ab\right)$

Therefore,

$2\log_3\left(8z\right)-\log_3\left(3^4-7^2\right)$

Applying power rule of logarithmic,

$a\log_c\left(b\right)=\log_c\left(b^a\right)$

Therefore,

$\log_3\left(\left(8z\right)^2\right)-\log_3\left(3^4-7^2\right)$

$\log_3\left(\left(64z^2\right)\right)-\log_3\left(3^4-7^2\right)$

Applying quotient rule of logarithmic,

$\log_c\left(a\right)-\log_c\left(b\right)=\log_c\left(\frac{a}{b}\right)$

Therefore,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{3^4-7^2}\right)$

Simplifying,

$\log_3\left(\dfrac{\left(64z^2\right)^2}{81-49}\right)$

$\log_3\left(\dfrac{\left(64z^2\right)^2}{32}\right)$

$\log_3\left(2z^2\right)$

Therefore value of expression is $\log_3\left(2z^2\right)$

6 0

4 years ago

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