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2 years ago

Anna is planning a birthday party for her brother Juan. She is planning on serving hamburgers to her guests.

1 answer:

7 0

The least amount of hamburger patties anna could buy would be 24 and that is 2 packs of 12. this is because the LCM of 12 and 8 is 24

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The number of bacteria in a certain culture doubled every hour. If there were 30 bacteria present in the culture initially, how

vichka [17]

Answer:

The number of bacteria after $8th$ will be $3840$

Step-by-step explanation:

Given the initially 30 bacteria present in the culture.

Also, the number of bacteria got doubled every hour.

So, using the equation

$A=A_0r^{n-1}$

Where $A$ is number of bacteria after $n$ hours.

$A_0$ is bacteria present initially.

$r$ is the common ration, in our problem it is given that bacteria doubles every hour. So, $r=2$

And $n$ is the number of hours. In our problem we need amount of bacteria at the end of $8th$ hours. So, $n=8$

Plugging values in the formula we get,

$A=30(2)^{8-1}\\A=30\times 2^7\\A=30\times 128\\A=3840$

So, number of bacteria after $8th$ will be $3840$

3 0

2 years ago

(Algebra II) Standard Form of a Quadratic Function - I'm having trouble with two questions on this quiz? Please don't just give

anastassius [24]

1.
the x value of the vertex in form
ax^2+bx+c=y
is
-b/2a
so

-2x^2+8x-18
x value of vertex is
-8/(2*-2)=-8/-4=2

plug it in to get y value
-2(2)^2+8(2)-18
-2(4)+16-18
-8-2
-10

vertex is at (2,-10)
or you could complete the square to get into y=a(x-h)^2+k, where the vertex is (h,k)
so as follows
y=(-2x^2+8x)-18
y=-2(x^2-4x)-18
y=-2(x^2-4x+4-4)-18
y=-2((x-2)^2-4)-18
y=-2(x-2)^2+8-18
y=-2(x-2)^2-10
vertex is (2,-10)

5.
vertex is the time where the speed is the highest
at about t=10, the speed is at its max

6 0

2 years ago

Read 2 more answers

What is the slope, and what is the y-intercept of the line?

Oliga [24]

$General\ equation\ for\ line\ in\ slope\ intercept\ form:
\\\\y=ax+b\\
 a-slope\\ 
b-intercept\\\\ 
y=\frac{5}{4}x+9\\\\
slope=\frac{5}{4}\\\\intercept=9$

4 0

2 years ago

Given two points M & N on the coordinate plane, find the slope of MN , and state the slope of the line perpendicular to MN .

telo118 [61]

Answer:

Problem 1) $m = \dfrac{1}{4}$ $slope_{perpendicular} = -4$

Problem 2) $m = \dfrac{1}{3}$ $slope_{perpendicular} = -3$

Step-by-step explanation:

$slope = m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$slope_{perpendicular} = \dfrac{-1}{m}$

Problem 1) M(9,6), N(1,4)

$slope = m = \dfrac{6 - 4}{9 - 1} = \dfrac{2}{8} = \dfrac{1}{4}$

$slope_{perpendicular} = \dfrac{-1}{\frac{1}{4}} = -4$

Problem 2) M(-2,2), N(4,-4)

$slope = m = \dfrac{4 - 2}{4 - (-2)} = \dfrac{2}{6} = \dfrac{1}{3}$

$slope_{perpendicular} = \dfrac{-1}{\frac{1}{3}} = -3$

4 0

2 years ago

Solve in radians help plz

PIT_PIT [208]

I think we can use the identity sin x/2 = sqrt [(1 - cos x) /2]

cos x - sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1

cos x - sqrt(3/2) sqrt(1 - cos x) = 1
sqrt(3/2)(sqrt(1 - cos x) = cos x - 1 Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1

cos^2 x - 0.5 cos x - 0.5 = 0

cos x = 1 , -0.5

giving x = 0 , 2pi, 2pi/3, 4pi/3 ( for 0 =< x <= 2pi)

because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.

The last 2 results dont fit so the answer is x = 0 , 2pi Answer

5 0

2 years ago