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3 years ago

Is a equilateral triangle always acute triangle

Mathematics

1 answer:

Vera_Pavlovna [14]3 years ago

3 0

Yes an equilateral triangle is a type or acute triangle.

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Find the degree of the monomial.<br> 3a4b3

steposvetlana [31]

Answer:

degree of the monomial is the sum of the exponents of all included variables

Step-by-step explanation:

please mark me as brainlist please

3 0

2 years ago

Solve 4(x + 2) = 20

KiRa [710]

Answer:

x = 3

Explanation:

$\sf 4(x + 2) = 20$

$\sf 4x + 8 = 20$

$\sf 4x = 20 - 8$

$\sf 4x = 12$

$\sf x = \frac{12}{4}$

$\sf x = 3$

check:

4(x + 2) = 20

4(3 + 2) = 20

4(5) = 20

20 = 20

Hence proved x = 3

4 0

2 years ago

Read 2 more answers

Gamal spent $12. 50 at the book store. The difference between the amount he spent at the video game store and the amount he spen

AysviL [449]

Answer: The options are kinda weird

But the solution to x - 12.5 = 17 is x = 29.5, so whichever equation you plug this into is correct is the right equation...

8 0

2 years ago

120 110 A pick-your-own blueberry farm offers 6 pounds of blueberries for $14. Cost (dollars) 100 90 Drag the points to create a

shutvik [7]

The relationship between cost and pounds of blueberries is y = (7/3)x

A linear equation is of the form:

y = mx + b

where y, x are variables, m is the slope of the line and b is the y intercept.

Let y represent the cost in dollars for x pounds of blueberries.

Since 6 pounds of blueberries cost $14, hence:

m = $14/ 6 pound = $7/3 per pound

Therefore the relationship between cost and pounds of blueberries is y = (7/3)x

Find out more at: brainly.com/question/25445678

5 0

3 years ago

A car is being driven at a rate of 40 ft/sec when the brakes are applied. The car decelerates at a constant rate of 10 ft/sec2.

IRISSAK [1]

Answer:

80 feet

Step-by-step explanation:

Given:

Initial speed of the car ( $v_0$ ) = 40 ft/sec

Deceleration of the car ( $\frac{dv}{dt}$ ) = -10 ft/sec²

Final speed of the car ( $v_x$ ) = 0 ft/sec

Let the distance traveled by the car be 'x' at any time 't'. Let 'v' be the velocity at any time 't'.

Now, deceleration means rate of decrease of velocity.

So, $\frac{dv}{dt}=-10\ ft/sec^2$

Negative sign means the velocity is decreasing with time.

Now, $\frac{dv}{dt}=\frac{dv}{dx}(\frac{dx}{dt})$ using chain rule of differentiation. Therefore,

$\frac{dv}{dx}\cdot\frac{dx}{dt}= -10\\\\But\ \frac{dx}{dt}=v.\ So,\\\\v\frac{dv}{dx}=-10\\\\vdv=-10dx$

Integrating both sides under the limit 40 to 0 for 'v' and 0 to 'x' for 'x'. This gives,

$\int\limits^0_{40} {v} \, dv=\int\limits^x_0 {-10} \, dx\\\\\left [ \frac{v^2}{2} \right ]_{40}^{0}=-10x\\\\-10x=\frac{0}{2}-\frac{1600}{2}\\\\10x=800\\\\x=\frac{800}{10}=80\ ft$

Therefore, the car travels a distance of 80 feet before stopping.

4 0

3 years ago