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2 years ago

One leg of an isosceles right triangle measures 5 inches. Rounded to the nearest tenth, what is the approximate length of

Mathematics

1 answer:

Firlakuza [10]2 years ago

6 0

The length of the hypotenuse = 7.1 inch

Step-by-step explanation:

In an isosceles right angle triangle if one leg is 5 inch, another side along the right angle is 5 in

hypotenuse = $\sqrt{side^{2} +base^{2} }$ ( by Pythagoras theorem)

= $\sqrt{5^{2}+5^{2} }$

= $\sqrt{50}$

= 7.07106781

= 7.1 inch (approx)

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Maya has 3 times as many dimes as nickels. She has 4 more quarters than nickels. if she has a total of $9.40, how many of each t

puteri [66]

Answer: nickels = 14, dimes = 42, quarters = 18

Step-by-step explanation:

Let n represent nickels which has a value of $0.05

Let d represent dimes which has a value of $0.10

Let q represent quarters which has a value of $0.25

It is given that:

n = n

d = 3n

q = n + 4

and their total value is $9.40 ⇒ n + d + q = $9.40

Use substitution to find the quantity of nickels:

.05(n) + .10(3n) + .25(n + 4) = 9.40

5n + 10(3n) + 25(n+4) = 940 multiplied by 100

5n + 30n + 25n + 100 = 940 distributed

60n + 100 = 940 added like terms

60n = 840 subtracted 100

n = 14 divided by 60

n = 14

d = 3n

= 3(14)

= 42

q = n + 4

= (14) + 4

= 18

4 0

2 years ago

Suppose a and b are the solutions to the quadratic equation 2x^2-3x-6=0. Find the value of (a+2)(b+2).

KonstantinChe [14]

Answer:

(a+2)(b+2) = 4

Step-by-step explanation:

We are given the following quadratic equation:

$2x^2-3x-6=0$

Let a a and b be the solution of the given quadratic equation.

Solving the equation:

$2x^2-3x-6=0\\\text{Using the quadratic formula}\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\text{Comparing the equation to }ax^2 + bx + c = 0\\\text{We have}\\a = 2\\b = -3\\c = -6\\x = \dfrac{3\pm \sqrt{9-4(2)(-6)}}{4} = \dfrac{3\pm \sqrt{57}}{4}\\\\a = \dfrac{3+\sqrt{57}}{4}, b = \dfrac{3-\sqrt{57}}{4}$

We have to find the value of (a+2)(b+2).

Putting the values:

$(a+2)(b+2)\\\\=\bigg(\dfrac{3+\sqrt{57}}{4}+2\bigg)\bigg(\dfrac{3-\sqrt{57}}{4}+2\bigg)\\\\=\bigg(\dfrac{11+\sqrt{57}}{4}\bigg)\bigg(\dfrac{11-\sqrt{57}}{4}\bigg)\\\\=\dfrac{121-57}{156} = \dfrac{64}{16} = 4$