Please help me 10 points

How would I figure out this answer<br>number 137, please help!!! due tomorrow

kaheart [24]

Who puts pencils in shoe boxes

5 0

3 years ago

The record high temperature recorded one year was 93°F while the record low temperature that year was -21°F. How much warmer was

gogolik [260]

Answer:

114°F

Step-by-step explanation:

114 + -23 = 93

3 0

3 years ago

Read 2 more answers

Jimrgrant or Someone Answer this PLZ

larisa [96]

Answer:

C and A

Step-by-step explanation:

For the triangle to be isosceles then the legs are congruent, that is

XZ = XY , substitute values

5x - 1 = 3x + 5 ( subtract 3x from both sides )

2x - 1 = 5 ( add 1 to both sides )

2x = 6 ( divide both sides by 2 )

x = 3 → C

(10)

The diagonals of a square bisect each other

Calculate PR using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2-y_{1})^2 }$

with (x₁, y₁ ) = P(0, 0) and (x₂, y₂ ) = R(6, - 8)

PR = $\sqrt{(6-0)^2}+(-8-0)^2$

= $\sqrt{6^2+(-8)^2}$

= $\sqrt{36+64}$

= $\sqrt{100}$

= 10

Then

PT = 0.5 × 10 = 5 → A

3 0

3 years ago

The graph below shows a quadratic function f(x) and an absolute value function g(x)

Sophie [7]

Answer:

The x-values that are approximate solution of the equation $f(x) = g(x)$ are $x_{1} \approx -3.5$ and $x_{2}\approx 5$ , respectively.

Step-by-step explanation:

From graph we infer that the x-values that represents a solution to the system of equations corresponds to those values of x in which both functions coincide. As we see, there are two x-values for $f(x) = g(x)$ , which are described below:

$x_{1} \approx -3.5$ and $x_{2}\approx 5$

The x-values that are approximate solution of the equation $f(x) = g(x)$ are $x_{1} \approx -3.5$ and $x_{2}\approx 5$ , respectively.

6 0

3 years ago

The daily production cost, C, for x units is modeled by the equation:<br> C = 200 – 7x + 0.345x2

lina2011 [118]

C(x) = 200 - 7x + 0.345x^2

Domain is the set of x-values (i.e. units produced) that are feasible. This is all the positive integer values + 0, in case that you only consider that can produce whole units.

Range is the set of possible results for c(x), i.e. possible costs.

You can derive this from the fact that c(x) is a parabole and you can draw it, for which you can find the vertex of the parabola, the roots, the y-intercept, the shape (it open upwards given that the cofficient of x^2 is positive). Also limit the costs to be positive.

You can substitute some values for x to help you, for example:

x y
0 200
1 200 -7 +0.345 = 193.345
2 200 - 14 + .345 (4) = 187.38
3 200 - 21 + .345(9) = 182.105
4 200 - 28 + .345(16) = 177.52
5 200 - 35 + 0.345(25) = 173.625
6 200 - 42 + 0.345(36) = 170.42

10 200 - 70 + 0.345(100) =164.5
11 200 - 77 + 0.345(121) = 164.745

The functions does not have real roots, then the costs never decrease to 0.

The function starts at c(x) = 200, decreases until the vertex, (x =10, c=164.5) and starts to increase.

Then the range goes to 164.5 to infinity, limited to the solutcion for x = positive integers.

8 0

3 years ago

Please help me 10 points​

Please help me 10 points