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Anna11 [10]
3 years ago
11

If anyone can help me with answering the pictures above i’d be really grateful

Mathematics
1 answer:
Usimov [2.4K]3 years ago
4 0
No it is not a perfect square
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Ted knows the actual distance between two cities is 150 miles. His map shows a distance of 2 inches between these cities. What i
Serjik [45]

Answer:

The scale used on his map is <u>150 miles : 2 inches</u>.

Step-by-step explanation:

Given:

Ted knows the actual distance between two cities is 150 miles. His map shows a distance of 2 inches between these cities.

Now, to find the scale Ted used on his map.

The actual distance Ted know between two cities = 150 miles.

The distance on map between these cities = 2 inches.

So, to get the scale used on his map:

\frac{The\ actual\ distance\ Ted\ know\ between\ two\ cities}{The\ distance\ on\ map\ between\ these\ cities}

=\frac{150\ miles}{2\ inches}

=150\ miles:2\ inches.

Therefore, the scale used on his map is 150 miles : 2 inches.

5 0
3 years ago
G(-4, -8) and H(-15, -3). What is the slope of GH?
Masteriza [31]

Answer:

G(x1,y1)= (-4,-8)

H(x2,y2)=(-15,-3)

Slope = (y2-y1)/(x2-x1)

=(-3-(-8))/ (-15-(-4))

= (-3+8)/(-15+4)

=5/-11

3 0
3 years ago
How can knowing the gcf and lcm help you when you add subtract and multiply fractions?
Maksim231197 [3]
<span>In order to add and subtract fractions, we must be able to find a <u>common denominator</u>.
The LCD, or least common denominator, is the LCM of the denominators. Once we've added, subtracted or multiplied fractions, we often have to simplify the fraction.
In order to do this, we divide the numerator and denominator by their GCF. This means we must be able to find the LCM and GCF in order to work with fractions.</span>
4 0
3 years ago
Read 2 more answers
Avani is building a rectangular play area. The length of the play area is 7.5 meters. The width of the play area is 5.3 meters.
const2013 [10]
39.8 is the answer blah blah
7 0
3 years ago
Read 2 more answers
If the right side of the equation dy dx = f(x, y) can be expressed as a function of the ratio y/x only, then the equation is sai
melomori [17]

Answer:

 y = x*sqrt(Cx - 1)

Step-by-step explanation:

Given:

                                  dy / dx = (x^2 + 5y^2) / 2xy

Find:

Solve the given ODE by using appropriate substitution.

Solution:

- Rewrite the given ODE:

                               dy/dx = 0.5(x/y) + 2.5(y/x)

- use substitution y = x*v(x)

                               dy/dx = v + x*dv/dx

- Combine the two equations:

                                v + x*dv/dx = 0.5*(1/v) + 2.5*v

                                x*dv/dx = 0.5*(1/v) + 1.5*v

                                x*dv/dx = (v^2 + 1) / 2v

-Separate variables:

                                 (2v.dv / (v^2 + 1) = dx / x  

- Integrate both sides:

                                 Ln (v^2 + 1) = Ln(x) + C

                                 v^2 + 1 = Cx

                                 v = sqrt(Cx - 1)

- Back substitution:

                                (y/x) = sqrt(Cx - 1)

                               y = x*sqrt(Cx - 1)

                         

3 0
3 years ago
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