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Umnica [9.8K]
3 years ago
11

1500(0.04)(3)=1500(0.04)(3)=

Mathematics
1 answer:
san4es73 [151]3 years ago
4 0
1500(0.04) = 60
60*3 or 60(3) = 180
Both equations are the same. I don't actually fully understand why you would use the same equation after the equal sign
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In 10 and 11, find the area in square units of each polygon.
alexandr1967 [171]

Answer:

10: 12 units^2 11: 33 units^2

Step-by-step explanation:

10:

Divide into square and 2 triangles (for simplicity)

square=9

one triangle=3/2

both triangles combined=3

9+3=12

11:

divide into 2 triangles

top triangle: height= 4, base=6

bottom triangle: height= 7, base=6

top=12

bottom=21

Add, 33.

7 0
2 years ago
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What is the exact value that is 59% of 2302?
serg [7]
59% of 2302....turn the percent to a decimal...." of " means multiply
0.59(2302) = 1358.18 <==
3 0
3 years ago
If f(x) = x^3 and g(x) = (x + 1)^3, which is the graph of g(x)?
STatiana [176]

Answer:

Step-by-step explanation:

the graph which passes through (-1,0)

5 0
3 years ago
I need help with my homework​
Gnom [1K]

Answer:

the whole picture isnt shown??

Step-by-step explanation:

6 0
3 years ago
Consider the solid S described below. The base of S is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sec
alisha [4.7K]

Answer:

Volume = \frac{64}{3}

Step-by-step explanation:

Given - Consider the solid S described below. The base of S is the triangular region with vertices (0, 0), (4, 0), and (0, 4). Cross-sections perpendicular to the x-axis are squares.

To find - Find the volume V of this solid.

Solution -

Given that,

The equation of the line with both x-intercept and y-intercept as 4 is -

\frac{x}{4} + \frac{y}{4}  = 1

⇒x + y = 4

⇒y = 4 - x

Now,

Volume = \int\limits^a_b {A(x)} \, dx

where

A(x) is the area of general cross-section.

It is given that,

Cross-sections perpendicular to the x-axis are squares.

So,

A(x) = (4 - x)²

As solid lies between x = 0 and x = 4

So,

The Volume becomes

Volume = \int\limits^4_0 {(4 - x)^{2} } \, dx

             = \int\limits^4_0 {[(4)^{2}  + (x)^{2} - 8x] } \, dx

             = \int\limits^4_0 {[16  + x^{2} - 8x] } \, dx

             = {[16 x  + \frac{x^{3}}{3}  - \frac{8x^{2} }{2} ] } ^4_0

             = {[16(4 - 0)  + \frac{4^{3}}{3} - \frac{0^{3}}{3}  - 4 [4^{2} - 0^{2}]   ] }

             = {[16(4)  + \frac{64}{3} - 0  - 4 [16 - 0]   ] }

             = {[64  + \frac{64}{3}  - 64  ] }

             = \frac{64}{3}

⇒Volume = \frac{64}{3}

7 0
3 years ago
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