Answer please........

Find the percent decrease round to the nearest percent from 93 Fahrenheit to 58 Fahrenheit the percent decrease is what percent

Elden [556K]

Answer:

38%

Step-by-step explanation:

Initial value = 93 F
Final value = 58 F

Decrease

93 - 58 = 35 F

Percent decrease

35/93*100% = 38 % rounded

5 0

3 years ago

PLEASE HELP 15 POINTS WILL GIVE BRAINLIEST The following formula, F = ma, relates three quantities: Force (F), mass (m), and acc

8_murik_8 [283]

Part A

The expression "ma" is the same as "m*a" (m times a). To isolate 'a', we divide both sides by m. Division is the inverse operation of multiplication. Think of it as undoing multiplication

F = ma
F = m*a
F/m = m*a/m
F/m = a
a = F/m

Answer: a = F/m
Note: the slash "/" without quotes means "divide"

===========================================
Part B

Use the result from part A. We will plug F = -24 and m = 10 into that equation

a = F/m
a = -24/10
a = -2.4

Answer: -2.4

===========================================
Part C

We will use the equation F = m*a

F = m*a
24 = m*12 ... plug in F = 24 and a = 12
24/12 = m*12/12
2 = m
m = 2

Answer: 2

4 0

3 years ago

You spent $10 on office supplies out of a $25 budget.What percent of your budget did you spend?

lana66690 [7]

You spent 40% of your budget.

10/25=2/5=40/100

6 0

3 years ago

a(x+b)=4x+10 in the equation above a and b are constants. if the equation has infinitely many solutions for X, what is the value

xeze [42]

Answer:

$b= \frac{5}{2}$

Step-by-step explanation:

Given

$a(x+b)=4x+10\\ax+b= 4x+10\\$

Now we know that system has infinite solution for x

$\frac{a_1}{a_2}=\frac{b_1}{b_2}$

in above equation.

$a_1=a, a_2=4, b_1=ab,b_2=10\\$

∴ $\frac{a}{4}=\frac{ab}{10}\\\therefore b=\frac{10\times a}{4\times a}=\frac{5}{2}$

7 0

3 years ago

Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4

steposvetlana [31]

$f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}$

$\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}$

6 0

4 years ago

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