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2 years ago

Which is the most likely first step needed to solve this equation?

Mathematics

2 answers:

Vanyuwa [196]2 years ago

5 0

Answer:

C. Add 15 to both sides

Step-by-step explanation:

Given equation:

99x – 15 = 45

To solve:

Step 1: Add 15 to both sides:

⇒ 99x – 15 + 15 = 45 + 15

⇒ 99x = 60

Step 2: Divide both sides by 99:

⇒ 99x ÷ 99 = 60 ÷ 99

⇒ x = 60/99

⇒ x = 20/33

erastovalidia [21]2 years ago

4 0

Answer:

Choice C: Add 15 to both sides.

Step-by-step explanation:

Given:

99x – 15 = 45

Aim:

To Find the first step.

Solution:

99x – 15 = 45

The first step will be Add 15 to both sides:

99x-15+15 = 45+15
99x + 0 = 65
99x = 60

Choice C is accurate.

$\rule{225pt}2pt$

EXTRA:

Let's solve the equation further.

Then transpose 99 to the RHS:

x = 60/99
x = 20/33

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Question 4 of 10

stiks02 [169]

Answer:

28° A

Step-by-step explanation:

we have isosceles triangle, which's 2 sides and 2 angles are equal.

we have two equal sides, so another side is base.

base angles are equal, so if E=28 thenG=28 too

3 0

3 years ago

Help please first with the right answer will be marked brainlyest

andre [41]

I believe this is correct

5 0

3 years ago

Read 2 more answers

Can someone help me please

djverab [1.8K]

Answer: 6 teams

Step by step:

3 people=1 group

3+3=6

6 people taken out of 18

2 groups already

3+6=9

9 out of 18 people taken

3 groups made

9+3=12

12 out of 18 people taken

4 groups made

12+3=15

15 out of 18 people taken

5 groups made

15+3=18

18 out of 18 people taken

Answer: 6 groups made

Short step by step:

18 divided by 3 equals 6

Or

3 times 6 equals 18

I really hope that made sense

8 0

2 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec

Learn more about the chain rule of differentiation here:

brainly.com/question/20433457

3 0

3 years ago

Evaluate the expression: 12-3x3-2 O 12 O9 O 25 0 1

nlexa [21]

Answer:

12-3×3-2

12-6-2

12-11

8 0

3 years ago