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2 years ago

Which expressions can you use to find 75% of 56? Check all that apply.

Mathematics

2 answers:

alisha [4.7K]2 years ago

7 0

Your answer would be B and C

son4ous [18]2 years ago

5 0

B and C
This is because 3/4 is the same as .75 and once you multiply them you are left with the correct answer causing it to be b and c!!

Hope this helps you!!

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The function f(x) = –10(x)(x – 4) represents the approximate height of a projectile launched from the ground into the air with a

sineoko [7]

The height of the project becomes zero when it hits the ground which is also the total time at which the projectile stays in the air. Equating the function to zero:
-10(x)(x - 4) = 0
The roots of this quadratic equation are:
x = 0, and x=4

Therefore, the projectile stays in the air for
4 seconds

4 0

2 years ago

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Jim’s family is choosing whether to go to a restaurant for lunch or for dinner. The cost of a salad is $7 at lunch and $9 at din

Vesna [10]

Answer:

To answer the problem above we let x be the number of people who prefers salad and y be the number of people who chose steak. For lunch they will be spending $41. That is,

7x + 10y = 41

Now, for dinner they will be spending $51. That is,

9x + 12y = 51

The values of x and y from both equations are 3 and 2, respectively. Therefore, 3 people will choose a salad

5 0

2 years ago

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. Solve the following initial value problem: (t2−20t+51)dydt=y (t2−20t+51)dydt=y with y(10)=1y(10)=1. (Find yy as a function of

Semenov [28]

Answer:

$y=(\frac{t-17}{t-3})^{\frac{1}{14}}$

Step-by-step explanation:

We are given that initial value problem

$t^2-20t+51)\frac{dy}{dt}=y$

$\frac{dy}{y}=\frac{dt}{t^2-20t+51}$

$\frac{dy}{y}=\frac{dt}{t^2-3t-17t+51}$

$\frac{dy}{y}=\frac{dt}{(t(t-3)-17(t-3)}$

$\frac{dy}{y}=\frac{dt}{(t-3)(t-17)}$

$\frac{1}{(t-3)(t-17)}=\frac{A}{t-3}+\frac{B}{t-17}$

$\frac{1}{(t-3)(t-17)}=\frac{A(t-17)+B(t-3)}{(t-3)(t-17)}$

$1=A(t-17)+B(t-3)$ ...(1)

Substitute t-3=0

t=3

t-17=0

t=17

Substitute t=3 in equation (1)

$1=A(3-17)+0$

$1=-14A$

$A=-\frac{1}{14}$

Substitute t=17

$1=B(17-3)$

$1=14B$

$B=\frac{1}{14}$

Substitute the values of A and B

$\frac{1}{(t-3)(t-17)}=-\frac{1}{14}(\frac{1}{t-3})+\frac{1}{14}(\frac{1}{t-17})$

$\int\frac{dy}{y}=-\frac{1}{14}\int\frac{dt}{t-3}+\frac{1}{14}\int\frac{dt}{t-17}$

$ln y=-\frac{1}{14}ln\mid{t-3}\mid+\frac{1}{14}\mid{t-17}\mid+ln C$

By using formula: $\frac{dx}{x}=ln x+C$

$ln y=\frac{1}{14}(-ln\mid{t-3}\mid+ln\mid{t-17}\mid)+ln C$

Using formula: $ln x-ln y=ln \frac{x}{y}$

$ln y=\frac{1}{14}(ln\mid{\frac{t-17}{t-3}}\mid)+ln C$

$ln y=\frac{1}{14}ln\mid{\frac{t-17}{t-3}}\mid+ ln C$

Substitute y(10)=1

$ln 1=\frac{1}{14}ln\mid\frac{10-17}{10-3}\mid+ln C$

$0=0+ln C$

Because ln 1=0

$lnC=0$

$C=e^0=1$

Because $ln x=y\implies x=e^y$

Substitute the value of C

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+ln1$

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid+0$

$ln y=\frac{1}{14}ln\mid\frac{t-17}{t-3}\mid$

$14ln y=ln\mid\frac{t-17}{t-3}\mid$

$lny^{14}=ln\mid\frac{t-17}{t-3}\mid$

By using identity $blog a= loga^b$

$y^{14}=\frac{t-17}{t-3}$

$y=(\frac{t-17}{t-3})^{\frac{1}{14}}$

6 0

2 years ago

Which expression is equivalent to 3 cube root (343x^9 y^12 z^6) ? 7x3y4z2 7x3y6z2 49x3y6z2 49x3y4z2

Andre45 [30]

Cube root of 343 = 7

of x^9 it is x^3 ( because x^9 = x^3 * x^3 * x^3)

of y^12 it is y^4

of z^6 it is z^2

So the correct choice is A

7 0

2 years ago

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Part A

Kitty [74]

Answer:

Flat fee of 20 dollars with 5 dollaers every hour=

20+h5

Step-by-step explanation:

8 0

2 years ago

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