MATHEMATICS HIGH SCHOOL

Markus scored 85, 92, 82, and 94 on the first four tests of the semester. His teacher has not yet told him his score on his fifth test, but did tell him that his score on the fifth test is five points lower than the average (arithmetic mean) of all five tests. Which equation could Markus use to determine the score of the fifth test, x?

Answers

Answer 1

Answer:

The answer is B. 5(x + 5) = 353 + x

(For anyone who needs it now)

Answer 2

Answer:

Answer: B. 5(x + 5) = 353 + x

Step-by-step explanation:

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Write down a number you can square to give an answer bigger than 900 but smaller than 1000

Answers

i think its 31 bcs
(31)² = 961
therefore,
900<961<1000

Which of the following best describes a three-dimensional solid made from a pile of similar circles that grow smaller as the solid gets taller until it reaches a point at the vertex?A . cylinder
B. cube
C. prism
D. cone

Answers

Your answer is D) Cone

Answer:

cone

Step-by-step explanation:

apeee

Receipt-of-goods discounts tend to be offered in cases where?

Answers

Answer:

Step-by-step explanation:

What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?f(x) = x4 – 3x2 – 4
f(x) = x3 + x2 – 4x – 4

Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2?
f(x) = –x2 – 8x – 16
f(x) = –x2 + 8x – 16
f(x) = –x2 – 8x + 16

Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1
f(x) = (x + 2i)(x + 3i)
f(x) = (x – 2)(x – 3)(x – 2i)(x – 3i)
f(x) = (x + 2i)(x + 3i)(x – 2i)(x – 3i)

Answers

In the question "What is the polynomial function of lowest degree with lead coefficient 1 and roots i, –2, and 2?" The given roots are i, -2 and 2. Recall that for any polynomial having complex root, the conjugate of the complex root is also a root of the polynomial, thus -i is also a root of the required equation. Thus the required equation is obtainrd thus: f(x) = (x - i)(x + i)(x - 2)(x + 2) = (x^2 + 1)(x^2 - 4) = x^4 - 4x^2 + x^2 - 4 = x^4 - 3x^2 - 4 In the question "Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2?" The required equation is obtained thus: f(x) = -(x - 4)^2 = -(x^2 - 8x + 16) = -x^2 + 8x - 16 In the question "Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1" Recall that for any polynomial having complex root, the conjugate of the complex root is also a root of the polynomial, thus -2i and -3i are also roots of the required equation. Thus the required equation is obtained thus: f(x) = (x + 2i)(x + 3i)(x - 2i)(x - 3i).

Solve 2 log2 2 + 2 log2 6 − log2 3x = 3.

Answers

Answer:

x = 6

Step-by-step explanation:

Given : $2\:log_2\:2\:+\:2\:log_26−\:log_2\:3x\:=\:3$

We have to solve the given expression $2\:log_2\:2\:+\:2\:log_26−\:log_2\:3x\:=\:3$

Subtract $2\log _2\left(2\right)+2\log _2\left(6\right)$ both sides , we have,

$2\:log_2\:2\:+\:2\:log_26-\:log_2\:3x-(2\log _2\left(2\right)+2\log _2\left(6\right)):=\:3-(2\log _2\left(2\right)+2\log _2\left(6\right))$

Simplify, we have,

$\log _2\left(3x\right)=3-2\log _2\left(2\right)-2\log _2\left(6\right)$

Divide both side by -1, we have,

$(-\log _2\left(3x\right))/(-1)=(3)/(-1)-(2\log _2\left(2\right))/(-1)-(2\log _2\left(6\right))/(-1)$

Simplify, we have,

$\log _2\left(3x\right)=-3+2\log _2\left(2\right)+2\log _2\left(6\right)$

Apply log rule, $a=\log _b\left(b^a\right)$

$2\log _2\left(6\right)-1=\log _2\left(2^(2\log _2\left(6\right)-1)\right)=\log _2\left(18\right)$

When log have same base,

$\log _b\left(f\left(x\right)\right)=\log _b\left(g\left(x\right)\right)\quad \Rightarrow \quad f\left(x\right)=g\left(x\right)$

$\mathrm{For\:}\log _2\left(3x\right)=\log _2\left(18\right)\mathrm{,\:\quad solve\:}3x=18$

3x = 18

x = 6

log(base2)[2² * 6² / 3x] = 3
144 / 3x = 2^3 = 8
144/8 = 3x
18 = 3x
x = 6

Algebraically solve the system of equations shown below. Note that you can use either factoring or the quadratic formula to find the X – coordinates, but the quadratic formula is probably easier.

Answers

$6x^2+19x-15=-12x+15\n6x^2+31x-30=0\n6x^2+36x-5x-30=0\n6x(x+6)-5(x+6)=0\n(6x-5)(x+6)=0\nx=(5)/(6) \vee x=-6\n\ny=-12\cdot(5)/(6) +15\vee y=-12\cdot(-6)+15\ny=-10 +15\vee y=72+15\ny=5 \vee y=87\n\nx=(5)/(6) \wedge y=5\nx=-6 \wedge y=87$

$y=6x^2+19x-15 \ny=-12x+15 \n \n6x^2+19x-15=-12x+15 \n6x^2+19x+12x-15-15=0 \n6x^2+31x-30=0 \n \n a=6 \n b=31 \n c=-30 \n \n x=(-b \pm √(b^2-4ac))/(2a)=(-31 \pm √(31^2-4 \cdot 6 \cdot (-30)))/(2 \cdot 6)=(-31 \pm √(1681))/(12)=(-31 \pm 41)/(12) \n x=(-31 -41)/(12) \ \hbox{or} \ x=(-31+41)/(12) \nx=-6 \ \hbox{or} \ x=(5)/(6)$

$y=-12 \cdot (-6)+15 \ \hbox{or} \ y=-12 \cdot (5)/(6)+15 \ny=87 \ \hbox{or} \ y=5 \n \n \hbox{the answer:} \n \boxed{x=-6, \ y=87} \ \hbox{or} \ \boxed{x=(5)/(6), \ y=5}$

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