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A square rug measures 8 feet by 8 feet. find the diagonal distance of the rug to the nearest whole number

Tanzania [10]

This is a right angle triangle and you use the Pythagorean theorem to solve for the hypotenuse (the diagonal)

A^2 + B^2 = C^2
8^2 + 8^2 = C^2
64 + 64 = C^2
128 = C^2
C = square_root (128)
C = 11 (to the nearest whole number)

The diagonal is 11 feet (to the nearest whole number)

7 0

3 years ago

If the domain of the function f(x) = x - 5 is {1, 3, 4, 6} what is the range? F

Margarita [4]

Answer:

-4, -2, -1, 1

Step-by-step explanation:

If the domain is given we use the equation to determine the range but usually the range would be everything in this equation.

7 0

3 years ago

State the value of the discriminate. Then determine the number of real roots of the equation.

nika2105 [10]

Answer:

no real solutions

Step-by-step explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 : a ≠ 0

Then the nature of the roots can be determined from the discriminant

b² - 4ac

• If b² - 4ac > 0 then 2 real and distinct roots

• If b² - 4ac = 0 then 2 real and equal roots

• If b² - 4ac < 0 then no real roots

Given

n(7n + 8) = - 10 ← distribute left side

7n² + 8n = - 10 ( add 10 to both sides )

7n² + 8n + 10 = 0 ← in standard form

with a = 7, b = 8 and c = 10, thus

b² - 4ac = 8² - (4 × 7 × 10) = 64 - 280 = - 216

Since b² - 4ac < 0 the equation has no real roots

8 0

3 years ago

PLEASE HELP (show work)

Talja [164]

Answer:

The book has 20 pages.

Step-by-step explanation:

100%-70%= 30%

6 pages = 30%

30%=6

:30 :30

1%=0,2

⋅100 ⋅100

100%=20

5 0

3 years ago

Read 2 more answers

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 10.4

vlabodo [156]

Answer:

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

Step-by-step explanation:

Bernoulli's equation,

$P_1+\frac12 \rho v^2_1+\rho g h_1= P_2+\frac12 \rho v^2_2+\rho g h_2$

P₁ = P₂= atmospheric presser

$\rho$ = density

$\frac12 \rho v^2_1+\rho g h_1= \frac12 \rho v^2_2+\rho g h_2$ [since P₁ = P₂]

$\Rightarrow\rho (\frac12 v^2_1+ g h_1)= \rho(\frac12 v^2_2+ g h_2)$

$\Rightarrow\frac12 v^2_1+ g h_1= \frac12 v^2_2+ g h_2$

$\Rightarrow\frac12 v^2_2-\frac12 v^2_1=g h_1- g h_2$

$\Rightarrow v^2_2- v^2_1=2g h$ [ $h_1-h_2=h$ ]

Here $v_1\approx 0$

$\Rightarrow v^2_2=2g h$

$\therefore v_2=\sqrt {2gh$

Here g= 9.8 m/s² , h = 10.4 m

The velocity of water that leaves from the hole $v_2$ = $\sqrt {2\times 9.8\times 10.4}$ m/s

=14.28 m/s.

Given, the rate of flow from the leak is $2.53\times 10^{-3}$ $m^3/min$

$=\frac{2.53\times 10^{-3}}{60}$ $m^3/s$

Let the diameter of the hole be d.

Then the cross section area of the hole is $=\pi (\frac d2)^2$

We know that,

The rate of flow = Cross section area × speed

$\Rightarrow \frac{2.53\times 10^{-3}}{60} =\pi (\frac d2)^2\times 14.28$

$\Rightarrow (\frac d2)^2=\frac{2.53\times 10^{-3}}{60\times 14.28\times \pi}$

$\Rightarrow d= 1.94 \times 10^{-3}$

Therefore the diameter of the hole is $1.94 \times 10^{-3}$ m.

4 0

3 years ago