What is 8/10 as a decimal

To lose one pound of fat, a 200-pound person must burn 3,500 calories. If that person burns 180 calories by walking for 30 minut

SVEN [57.7K]

Answer:

C - 19 hours 27 minutes

Step-by-step explanation:

180 × 2 = 360

360 calories in an hour

3500 × 2 = 7000 (amount of calories in 2 pounds)

7000 ÷ 360 = 19.4444...

I hope this helps

4 0

3 years ago

Alina drew a model of a square pyramid. The dimensions of the model are shown in the diagram. pls don't answer with links

Romashka [77]

Answer:

C

Step-by-step explanation:

Volume of a pyramid = 1/3 base area * height

= 1/3 (8 * 10)(12)

6 0

3 years ago

Which group of numbers includes only composite numbers?

Oksi-84 [34.3K]

Answer: 4, 6, 8, 9 and 10

Why?: What is the only composite number?

A composite number is a positive integer. which is not prime (i.e., which has factors other than 1 and itself)

8 0

3 years ago

Find f(g(x)). State the domain of the composite function.<br> f(x) = 3x-2/x+1<br> g(x) = x+5/2x-3

AnnyKZ [126]

Answer:

ℝ - {(-2/3),(3/2)}

Step-by-step explanation:

We want the domain of f(g(x)). So, firstly, we have to find the domain for g(x) and, then, for f(g(x)).

- Domain of g(x): Since the expression is a fracion, we must exclude the values of x that make null the denominator. Hence,

$g(x)=\dfrac{x+5}{2x-3}\Longrightarrow 2x-3\neq 0\iff \boxed{x\neq\dfrac{3}{2}}$

- Domain of f(g(x)): We'll find its expression:

$f(x) = \dfrac{3x-2}{x+1}\\\\f(g(x)) = \dfrac{3g(x)-2}{g(x)+1}\\\\f(g(x)) = \dfrac{3\cdot\dfrac{x+5}{2x-3}-2}{\dfrac{x+5}{2x-3}+1}=\dfrac{~~~\dfrac{3(x+5)-2(2x-3)}{2x-3}~~~}{\dfrac{(x+5)+(2x-3)}{2x-3}}\\\\f(g(x)) =\dfrac{3(x+5)-2(2x-3)}{(x+5)+(2x-3)}=\dfrac{3x+15-4x+6}{x+5+2x-3}\\\\\boxed{f(g(x)) =\dfrac{21-x}{3x+2}}$

Now, once again, we have to exclude the values of x that make the denominator equals to zero. Thus,

$f(g(x)) =\dfrac{21-x}{3x+2}\Longrightarrow 3x+2\neq0\iff \boxed{x\neq-\dfrac{2}{3}}$

Lastly, we may write the domanin of f(g(x)):

$D(f(g(x)) = \left]-\infty,-\dfrac{2}{3}\right[\cup\left]-\dfrac{2}{3},\dfrac{3}{2}\right[\cup\left]\dfrac{3}{2},\infty\right[$

or, just writing in a shorter way:

$\boxed{D(f(g(x)) = \mathbb{R}-\left\{-\dfrac{2}{3},\dfrac{3}{2}\right\}}$

7 0

3 years ago

There are 150 marigold plants in a back yard. Each month, the number of marigold plants decreases by 15%. There are 125 sunflowe

DENIUS [597]

To solve this problem, let us first assign variables. Let us say that:

X = number of marigold plants

Y = number of sunflower plants

n = number of months

We can see that in the given problem, X is decreasing by a percentage, this means that we have to set-up a geometric equation while for Y the decrease is linear so we set-up an arithmetic equation.

Part A.

For marigold plants X, a geometric sequence has a general form of:

X = Xo * (1 + r)^n

where r = -15% = -0.15 (negative since it is decreasing)

Xo = the initial amount of marigold plants = 150

X = 150 * (1 – 0.15)^n

X = 150 (0.85)^n

For the sunflower plants Y, an arithmetic sequence has a general form of:

Y = Yo + d * n

where d = -8 and Yo = 125

Y = 125 – 8 n

Part B. For n = 3

X = 150 (0.85)^3 = 92.12 = 92

Y = 125 – 8 (3) = 101

Part C. From Part B we see that the two values are very far from each other when n = 3, therefore they must be similar when n < 3. So we try n = 2

X = 150 (0.85)^2 = 108.38 = 108

Y = 125 – 8 (2) = 109

Therefore the two plants have approximately similar amount after 2 months.

6 0

3 years ago