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2 years ago

13

The altitude of a triangle is increasing at a rate of 1 cmymin while the area of the triangle is increasing at a rate of 2 cm2ym

in. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2

1 answer:

mr_godi [17]2 years ago

7 0

Answer:

b

Step-by-step explanation:

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Which expression has the greatest value when x=3

Luden [163]

Answer:

The third expression: (2x)^3

Step-by-step explanation:

Plug in 3 for x in each expression

2x^3

2(3)^3

Use PEMDAS

2 (3 × 3 × 3)

2 (9 × 3)

2 (27)

54

2x^3 + 5

(The last expression told us 2x^3 = 54 so feel free to use this information to make life easier)

54 + 5

59

(2x)^3

(2 × 3)^3

Use PEMDAS

6^3

6 × 6 × 6

36 × 6

216

(x - 1)^3

(3 - 1)^3

Use PEMDAS

2^3

2 × 2 × 2

4 × 2

8

216 is the greatest value so the third expression is the answer

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2 years ago

Assume that in October of a particular year about 215 billion text messages were sent or received in the US, and the prediction

Julli [10]

Answer:

There will be $243$ billion messages sent or received in November.

Step-by-step explanation:

Given.

Number of text messages sent or received $=215$ billion.

Increment on the very next month $=13\%=\frac{13}{100}=0.13$

Lets find what is $13\%\ of\ 215$

⇒ $13\%\ of\ 215$

⇒ $0.13\times 215=27.95$

So we will add this value to our previous months data.

Now

Number of text messages sent in the month of November $=(27.5+215)=242.95$ billion.

Rounding to the nearest tenth the answer will be $243$ billion.

So $243$ billion text messages were sent or received in November.

5 0

2 years ago

What is this number in standard form? plzzzzzzzzz i need this

Nookie1986 [14]

The answer is 82.046

3 0

2 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

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2 years ago

Write an absolute value equation that satisfies the following condition.

lawyer [7]

Answer:

x=-17.......................

Step-by-step explanation:

x+17=0

3 0

2 years ago