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balu736 [363]
2 years ago
7

HELP ASAP 10 POINTSSSS

Mathematics
1 answer:
lord [1]2 years ago
6 0

Use the Pythagorean theorem   (a^2 + b^2 = c^2)

7^2 + 24^2 = x^2

x^2 = 625

x = 25

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On the grid draw the graph of x + 2y =7
Elena-2011 [213]

Answer:

See attachment for graph

Step-by-step explanation:

Given

x + 2y = 7

Required

Draw the graph

First, we identify at least 2 possible values of x and y

Let x = 1

Solve for y

x + 2y = 7

1 + 2y = 7

2y = 6

y = 3

So, a point is: (1,3)

Let x = 3

Solve for y

x + 2y = 7

3 + 2y = 7

2y = 7 -3

2y = 4

y = 2

So, another point is: (3,2)

<em>Plot </em>(1,3)<em> and </em>(3,2)<em> on the graph, then draw a straight line through both points</em>

See attachment for graph

5 0
3 years ago
Pedaler's peddlers is a bike store that is having there annual sale. Denise wants to buy a bike that originally cost $264. The b
ANEK [815]
105.6

Convert 40% to 0.4
Then multiply 0.4 by 264
0.4*264= 105.6
5 0
2 years ago
Circumstance of 12yd
marusya05 [52]

Answer: 40.82yd

Step-by-step explanation:

First of all, it's ''circumference'' and second of all, it's ''13yd''.

Formula: C=2\pi r

2r is the same as d = diameter,

C=\pi d

C=(3.14)(13)\\C=40.82yd

5 0
3 years ago
Read 2 more answers
Solve the system: 2x+3y=6 and x=-2y+3. I need help
Alex Ar [27]

Answer:

\large\boxed{x=3,\ y=0\to(3,\ 0)}

Step-by-step explanation:

\left\{\begin{array}{ccc}2x+3y=6&(1)\\x=-2y+3&(2)\end{array}\right\\\\\text{Substitute (2) to (1):}\\\\2(-2y+3)+3y=6\qquad\text{use distributive property}\\\\(2)(-2y)+(2)(3)+3y=6\\\\-4y+6+3y=6\qquad\text{subtract 6 from both sides}\\\\-y=0\to\boxed{y=0}\\\\\text{Put the value of y to (2):}\\\\x=-2(0)+3\\\\x=0+3\\\\\boxed{x=3}

3 0
3 years ago
a car was valued at $41,000 in the year 2009 by 2013 the car value has depreciated to 19,000 if the car value continues to by th
lubasha [3.4K]

Answer:

$6,376.92

Step-by-step explanation:

-Let d be the rate of depreciation per year.

-Therefore, the value after n years can be expressed as:

A=P(1-d)^n\\\\A=Value \ after \ n  \ years\\P=Initial \ Value\\d=Rate \ of \ depreciation\\n=Time \ in \ years

#We substitute for the years 2009-2013 to solve for d:

A=P(1-d)^n\\\\19000=41000(1-d)^4\\\\0.475=(1-d)^4\\\\d=1-0.475^{0.25}\\\\d=0.1698

#We then use the calculated depreciation rate above to solve for A after 10 yrs:

A=P(1-d)^n\\\\=41000(1-0.1698)^{10}\\\\=\$6,376.92

Hence, the value of the car after 10 yrs is $6,376.92

3 0
3 years ago
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