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2 years ago

40% of students have brown eyes 10 have brown eyes how many atudents in the class

Mathematics

2 answers:

maxonik [38]2 years ago

6 0

50 I believe I think that 40% is 40 + 10 = 50?

castortr0y [4]2 years ago

3 0

Answer:

Step-by-step explanation:

40%x = 10

40/100 x=10

x=10* 100/40

x = 25

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The function is given by the formula f(x)=3x+2. a Find f(7).

Artyom0805 [142]

Answer:

Step-by-step explanation:

plug in 7

f(x)=3(7)+2

multiply 3 and 7 which is 21

and 2

f(x)=23.

5 0

3 years ago

Which angles are vertical angles?

lesya692 [45]

Answer:

<DCA and <BCF

Step-by-step explanation:

vertical angles are the opposite angles created by intersecting lines and are congruent. Vertical angles share only a vertex

3 0

3 years ago

X²- 12x + a = (x-b)² <br>Find the Value of a and value of b <br>

maxonik [38]

Answer:

a = 36 , b = 6

Step-by-step explanation:

expand the right side using FOIL and compare coefficients with like terms on left side.

(x - b)²

= x² - 2bx + b² , compare to x² - 12x + a , then

- 2b = - 12 ( divide both sides by - 2 )

b = 6

and a = b² = 6² = 36

5 0

1 year ago

Can you please please help me in this Math Question!!!!! Be sure to Show Your Work/Steps!!!!!

noname [10]

Just divide the original cost for one pound by 4 for each meat.
4.25 divided by 4. if it makes it any easier, 425 divided by 400 because you just move the decimal over and you will get the same result.
2.85 (or 285) divided by 4 (or 400).
then add the original cost per pound for each and then subtract turkey from beef.

8 0

3 years ago

Integration by Parts Evaluate e-2x cos(2x) dx.

kifflom [539]

Let

$I = \displaystyle \int e^{-2x} \cos(2x) \, dx[/]texIntegrate by parts:[tex]\displaystyle \int u \, dv = uv - \int v \, du$

with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \cos(2x) \, dx \implies v = \dfrac12 \sin(2x)$

Then

$\displaystyle I = \frac12 e^{-2x} \sin(2x) + \int e^{-2x} \sin(2x) \, dx + C$

Integrate by parts again, this time with

$u = e^{-2x} \implies du = -2 e^{-2x} \, dx \\\\ dv = \sin(2x) \, dx \implies v = -\dfrac12 \cos(2x)$

so that

$\displaystyle I = \frac12 e^{-2x} \sin(2x) - \frac12 e^{-2x} \cos(2x) - \int e^{-2x} \cos(2x) \, dx + C\\\\ \implies I = \frac{\sin(2x)-\cos(2x)}{2e^{2x}} - I + C \\\\ \implies 2I = \frac{\sin(2x) - \cos(2x)}{2e^{2x}} + C \\\\ \implies I = \boxed{\frac{\sin(2x) - \cos(2x)}{4e^{2x}} + C}$

6 0

1 year ago