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2 years ago

6

Members of the Science Club went to a history museum. It cost $7.25 for each member of the club. If 90 members went to the museu

m, how much would the total cost be?

1 answer:

malfutka [58]2 years ago

5 0

652 I believe or something close to that.

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The phone company A Fee and Fee has a monthly cellular plan where a customer pays a flat

11111nata11111 [884]

Answer:

<u>The equation is:</u>

y = mx + b

<u>We have two points:</u>

(310, 68) and (620, 99)

A)

<u>Find the slope using these two points:</u>

m = (99 - 68)/(620 - 310) = 31/310 = 0.1

<u>Use one of the points to find the y-intercept b:</u>

68 = 0.1*310 + b
68 = 31 + b
b = 68 - 31
b = 37

<u>The equation is:</u>

y = 0.1x + 37

B)

<u>Find the value of y when x = 980:</u>

y = 0.1*980 + 37
y = 98 + 37
y = 135

The total cost is $135

8 0

2 years ago

What's the length of a segment that begins at the point (2, 3) and ends at the point (18, 15)?

DerKrebs [107]

So first of all we need to find the slope and it will give us the answer what is the length of a segment.

To find it we are using the formula:
y2-y1/x2-x1.

15-3/18-2=12/16=3/4.
So I think that the length of the segment is 3/4 cm or what you didn’t tell it for me. Hope this helps.

6 0

3 years ago

masha68 [24]

Multiply the area of all sides (10*5)*2+(10*4)*2+(4*5)*2 to get 220 in^2

5 0

2 years ago

Evaluate the expression 62 + x-x2 for x=3

AVprozaik [17]

Answer:

56

Step by step explanation:

$62+x-x^2\\\\\text{When x = 3}\\\\62+3 -3^2 = 62+3-9 = 65-9 = 56$

8 0

1 year ago

In rectangle WXYZ, A is on side WX such that AX = 4, B is on side YZ such that BY = 18, and C is on side XY such that angle ACB=

Lemur [1.5K]

Answer:2*sqrt(130)

Step-by-step explanation: $From right triangle $AXC$, we have $\angle XAC = 90^\circ - \angle XCA$. We also must have $\angle YCB + \angle ACB + \angle XCA = 180^\circ$, so $\angle YCB = 180^\circ - 90^\circ - \angle XCA = 90^\circ - \angle XCA$, which means $\angle YCB = \angle XAC$. Combining this with $\angle X = \angle Y$, we have $\triangle XCA \sim \triangle YBC$ by AA Similarity, and\[\frac{CX}{AX} = \frac{BY}{CY},\]so\[\frac{CX}{4} = \frac{18}{2CX}.\]\\\\This gives us $2CX^2 = 72$, so $CX^2 =36$ and $CX=6$. \\\\\\\\$ $Therefore, $CY = 12$.Applying the Pythagorean Theorem to right triangles $XCA$ and $CYB$ gives us\begin{align*}CA^2 &= XA^2 + XC^2 = 16 + 36 = 52,\\BC^2 &= CY^2 + BY^2 = 144 + 324 = 468.\end{align*}Applying the Pythagorean Theorem to right triangle $ABC$ gives\[AB = \sqrt{AC^2 + BC^2} = \sqrt{52 + 468} = \sqrt{520} = \boxed{2\sqrt{130}}.\](We can also compute $AB$ by dropping a perpendicular from $A$ to $\overline{YZ}$, which creates a right triangle.)$

3 0

3 years ago