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3 years ago

A large grouper can weigh 1/3 ton. how much doe a large grouper weigh to the nearest pound?

Mathematics

1 answer:

zvonat [6]3 years ago

7 0

Well, 1 ton = 2000 IBS

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Choose the correct equation for the function of the graph below.

Wittaler [7]

Answer:

Y= 1/2 cos4X (Option 3)

Step-by-step explanation:

It cuts axis at 0.5

8 0

3 years ago

11 The sides of two similar triangle are in a ratio of 5:6. The area of the larger triangle is 108.

Elodia [21]

Answer:

The area of the smaller triangle is 75.

Step-by-step explanation:

Here, the ratio of the sides of the similar triangle are 5 : 6

The area if the larger triangle = 108

let us assume that the area of the smaller triangle = m

By the Theorem:

In two similar triangles, the ratio of the areas of similar triangles is the square of the ratio of their sides.

Similarly, here

$[tex](\frac{5}{6}) ^2$ = \frac{m}{108}[/tex]

⇒ $\frac{(5)^2}{(6)^2} = \frac{m}{108}$

or, $\frac{25}{36} = \frac{m}{108} \implies m = \frac{108 \times 25}{36} = 75$

⇒ m = 75

Hence, the area of the smaller triangle is 75.

6 0

3 years ago

Combine the following sets of terms.

DanielleElmas [232]

Let's simplify step-by-step To understand :)

3x2−7x−(x2+3x−9)

Distribute the Negative Sign:

=3x2−7x+−1(x2+3x−9)

=3x2+−7x+−1x2+−1(3x)+(−1)(−9)

=3x2+−7x+−x2+−3x+9

Combine Like Terms:)

=3x2+−7x+−x2+−3x+9

=(3x2+−x2)+(−7x+−3x)+(9)

=2x2+−10x+9

Answer :)

=2x2−10x+9

HOPE THIS HELPS

3 0

3 years ago

A model of a famous statue is 2 1/2 inches tall. The actual statue is 3 1/3 feet tall. What is the ratio of the height of the mo

Gekata [30.6K]

1. A model of a famous statue is $2\dfrac{1}{2}$ inches tall that is

$\dfrac{2\cdot 2+1}{2}=\dfrac{5}{2}$ in.

2. The actual statue is $3\dfrac{1}{3}$ feet tall that is

$\dfrac{3\cdot 3+1}{3}=\dfrac{10}{3}\ ft=\dfrac{10}{3}\cdot 12=40\ in.$

3. The ratio of the height of the model to the height of the actual statue in simplest form is

$\dfrac{\dfrac{5}{2}}{40}=\dfrac{5}{2}\cdot \dfrac{1}{40}=\dfrac{1}{16}.$

Answer: $\dfrac{1}{16}.$

4 0

3 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

3 years ago