1. 1 71 What is the solution to the equation 74-3=7+7X? X=-5 X=-4 X=4 X=5

<img src="https://tex.z-dn.net/?f=%28v%2B6%29%5E%7B2%7D%3D2v%5E%7B2%7D%2B14v%2B12" id="TexFormula1" title="(v+6)^{2}=2v^{2}+14v+

timofeeve [1]

Answer:

v=-6 or 4

Step-by-step explanation:

3 0

3 years ago

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Select the correct answer.

vovikov84 [41]

(Remember, PEMDAS)
First find what 5^2 equal to
5 • 5 = 25
Then add 25 to 3
25 + 3 = 28
So Oc. :)

7 0

3 years ago

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

6 0

3 years ago

A circular pool with a diameter of 18 ft will have a uniform 4 ft concrete walkway poured around it. If the concrete cost $4.25

zysi [14]

A circular pool with a diameter of 18 ft will have a uniform 4 ft concrete walkway poured around it. If the concrete cost $4.25 a square foot, how much will it cost for the concrete?

This can be solve be solving the area of walkway

Area of walkway is equal to = (pi)( 9 ft +4) ^2– (pi)(9ft)^2

= 276.46 sq ft

Cost = 276.46 sq ft * ($4.25 a square foot) = $1174.70

4 0

3 years ago

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The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func

alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

$h(t)=-16t^2 + 96t$

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

$To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)$

$Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution$

Conclusion:

The domain of the function is the time in between 0 to 6 seconds

$0 \leq t \leq 6$

The height will be positive in the above interval.

7 0

3 years ago

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