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mixas84 [53]
3 years ago
5

PLEASE HELP ASAP:

Mathematics
2 answers:
dmitriy555 [2]3 years ago
8 0

Answer:

39.34 juniors and 30.1 seniors

Slav-nsk [51]3 years ago
6 0

Answer:

The closest number of juniors and seniors who were inducted is 69.

Step-by-step explanation:

At Lincoln High School, approximately 7% of enrolled juniors and 5% of enrolled seniors were inducted into the National Honor Society last year.

There were 562 juniors and 602 seniors enrolled at Lincoln High School last year.

We will calculate the number of students as:

0.07(562)+0.05(602)

= 39.34+30.1

= 69.44

Therefore, the closest number of juniors and seniors who were inducted is 69.

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Factor the expression. x^2 – x – 42
IgorC [24]
X^2 - x - 42 = x^2 - 6x + 7x - 42 = x( x - 6) + 7(x - 6) = (x-6)(x+7);
The correct answer is a.
5 0
3 years ago
Can anyone help me on this
aliya0001 [1]

Answer:

-6 and 1

Step-by-step explanation:

If you separate the different powers (^), you have -m^3 - 5m^3 and -m^2 + 2m^2.

-m^3 - 5m^3 = -6m^3

-m^2 + 2m^2 = m^2 or 1m^2

4 0
3 years ago
The box-and-wishker plots bellow
Natasha2012 [34]

Answer:

H: Plots 1 & 2

Step-by-step explanation:

4 0
2 years ago
X is inversely proportional to y. X is directly proportional to z squared. When y = 4, x = 5.5. Work out the value of x when y =
Hitman42 [59]

4.4

Step-by-step explanation:

X is inversely proportional to y, so x = k/y where k is just a constant.

X is directly proportional to z squared, so x = kz^2 where k is a constant

Here, The second relationship doesn't even matter.

Plug the values for x and y into the first equation and solve for k.

5.5=k/4

k=5.5*4

k=22

Now plug k and the second y back in.

x=22/5

x=4.4

3 0
3 years ago
the length of each side of the ABCD EFGH cube is 6cm. If point P is located in the middle of line EH, point Q is in the middle o
azamat

Answer:

           The distance is:   \sqrt3\ cm\approx1,73\,cm

Step-by-step explanation:

The distance of point E to the PQR plane it is the hight (vertical) of piramid PRQE

If point P is located in the middle of line EH, point Q is in the middle of line EF, and point R is in the middle of line AE than:

EP = EQ = ER = 0.5EF = 3 cm  and  m∠REQ = m∠QEP = m∠REP = 90° so triangles RQE, QPE and PRE are congruent.

RQ = QP = PR so triangle PQR is equilateral and from Pythagorean theorem (for ΔRQE):

RQ^2=ER^2+EQ^2=3^2+3^2=2\cdot3^2\ \ \implies\ \ RQ=3\sqrt2

Then: RN=\dfrac{RQ\,\sqrt3}2

and:  RK=\dfrac23RN=\dfrac{RQ\,\sqrt3}3=\dfrac{3\sqrt2\cdot\,\sqrt3}3=\sqrt6

Therefore from Pythagorean theorem (for ΔERK):

EK^2+RK^2=ER^2\\\\EK^2=ER^2-RK^2\\\\EK^2=3^2-(\sqrt6)^2\\\\EK^2=9-6=3\\\\EK=\sqrt3\ cm\approx1,73\,cm

8 0
3 years ago
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