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tamaranim1 [39]
3 years ago
10

What is the sum of 2 3/5 - 1 3/8

Mathematics
1 answer:
Natalija [7]3 years ago
8 0
The Sum of 2 3/5-1 3/8 is 1.225
i hoped that helped u. 
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If the sun is 55° above the horizon, find the length of the shadow cast by a building 88 ft tall. Round your answer to the neare
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Answer: 61.6

Step-by-step explanation:

Use a calculator if you search it up and plug in the number it’ll give you the answers

7 0
2 years ago
A line with a slope of -4 has a coordinate of (0,3). If the other coordinate is (6,t),
s2008m [1.1K]

Answer:

Alrighty! So this is what we know:

Slope (m) = (-4)

Co-ordinate (0,3)

Co-ordinate 2 (6,t)

t = ?

Let's find our equation...

y=mx+b <-- just plug what we know in...

3=(-4)0+b

3=0+b

b=3

We just found b! woohoo

Now what is t? t is a value we need to know WHICH IS the y value of a co-ordinate that corresponds to when x is 6, as we know a co-ordinate is expressed as --> (x value, y value).

SO-

t=y

plug it in our equation...

y=mx+b

t=(-4)x+3

t=(-4)(6)+3

t=(-24)+3

t=(-21)

SOOOOOOOO WE HAVE OUR T!! just write a therefore statement like "therefore the value of t is -21." and you're set! hope this helped, any questions? ask in the comment section. :)

Step-by-step explanation:

7 0
2 years ago
Can y’all help me please !
Leviafan [203]

no because she turned it only 90 degrees not 180.

180 would be a mirror reflection

6 0
3 years ago
The length of a rectangle is 10cm longer than its breadth the perimeter is 48cm determine the area of the rectangle​
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5 0
2 years ago
Read 2 more answers
On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

5 0
3 years ago
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