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3 years ago

How much is 3/10 times 7

Mathematics

2 answers:

Fudgin [204]3 years ago

7 0

Answer:

Step-by-step explanation:

Rashid [163]3 years ago

4 0

3/10 x 7 = 21/10
or alternative forms: 2 1/10, 2.1

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You start a lawn care company with four friends. You agree to split the profits equally. The tools you need to buy cost $525. Ho

elena55 [62]

Because you're starting the company with 4 friends, there are 5 people in total. Therefore, if you do (400 × 5) + 525, you get the answer. 400 × 5 = 2000, and 2000 + 525 = $2525. I hope this helps! Let me know if you need me to explain anything :)

6 0

3 years ago

Last week, it rained g inches. This week, the amount of rain decreased by 5%. Which expressions represent the amount of rain tha

dybincka [34]

Answer:

The correct answers are:

0.95g

g - 0.05g

(1 - 0.05)g

Step-by-step explanation:

Amount of last week's rain = g inches

this week = 5% decrease on last week's amount

5% of last week's amount = 5/100 × g = 0.05g

∴ This week's amount of rain = 0.05g decrease

∴ Amount of rain this week = (Amount of rain last week) - 0.05g

∴ Amount of rain this week = g - 0.05g

This can be simplified by factorizing the common term "g"

g - 0.05g = g(1 - 0.05) = 0.95g

4 0

3 years ago

If s(x)=x-7 and t(x) =4x^2+3 which expression is equivalent to (t×s)(x)

marshall27 [118]

Answer:

$\large\boxed{(t\times s)(x)=4x^3-28x^2+3x-21}$

Step-by-step explanation:

$s(x)=x-7\\t(x)=4x^2+3\\\\(t\times s)(x)=t(x)\times s(x)\\\\\text{Substitute:}\\\\(t\times s)(x)=(4x^2+3)(x-7)\\\\\text{use FOIL:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(t\times s)(x)=(4x^2)(x)+(4x^2)(-7)+(3)(x)+(3)(-7)\\\\=4x^3-28x^2+3x-21$

8 0

3 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .