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3 years ago

On a map the scale is given as 2.5 inch: 4 miles. How many miles in real life is 5 inches on the map? *

Mathematics

1 answer:

liubo4ka [24]3 years ago

5 0

Answer:

8 miles

Step-by-step explanation:

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Combining Like Terms

Vsevolod [243]

I'm a bit confused.. Is this supposed to be asking a question about Combining Like Terms or is it like, telling what it is or something. I'm just clueless lol, It's good if your just telling people what it is if anyone needs help with it though! Not many people do that!

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3 years ago

(5 × 4) × (16 ÷ 8) × (24 − 22) =

sweet [91]

The answer should be 80

8 0

3 years ago

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Find the value of x that will make A||B. A B 4x 2x X = ?

pashok25 [27]

Answer:

x = 30

Step-by-step explanation:

Lines A and B are parallel lines and a transverse line is intersecting these lines at two distinct points.

By the definition of interior consecutive angles, sum of the measures of interior consecutive angles is 180°.

4x + 2x = 180°

6x = 180

x = 30

Therefore x = 30 will be the answer.

8 0

3 years ago

A coordinate grid with 2 lines. The first line is labeled y equals negative StartFraction 7 over 4 EndFraction x plus StartFract

KengaRu [80]

Answer:

1) (2.2, -1.4)

2) (1.33, 1)

Step-by-step explanation:

Question 1)

Two lines, with their corresponding equations are given and we have to find the solution to the system of equations.

The given lines are:

Equation of Line 1:

$y=\frac{-7}{4}x+\frac{5}{2}$

This line passes through the points: (0, 2.5) , (2.2, -1.4)

Equation of Line 2:

$y=\frac{3}{4}x-3$

This line passes through the points (0, -3) , (2.2, -1.4)

By looking at the graph/given data we have to find the solution of these linear equations.

Remember that the solution of linear equations is an ordered pair, through which both the lines pass i.e. the point at which both the given lines intersect is the solution of the linear equations.

From the given data we can see that both the lines pass through one common point, (2.2, -1.4). Since, both lines pass through this point, this means this is the point of intersection of the lines and hence there solution.

So, the answer to this questions is (2.2, -1.4)

Question 2)

The given equations are:

y = 1.5x - 1 Equation 1

y = 1 Equation 2

We can solve these equations by method of substitution.

Substituting the value of y from Equation 2, in Equation 1, we get:

1 = 1.5x - 1

1 + 1 = 1.5x

2 = 1.5x

x = 2/1.5

x = 1.33

y = 1

Thus, the solution of the given linear equations is (1.33, 1)

5 0

3 years ago

Read 2 more answers

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

1 year ago