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2 years ago

A biker rides for h hours at a speed of s miles per hour and travels d miles. Which equation models the relationship?

Mathematics

2 answers:

Luda [366]2 years ago

7 0

I believe it’s B good luck!!

Dmitry_Shevchenko [17]2 years ago

6 0

Answer:

D=SH so it would be B

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<img src="https://tex.z-dn.net/?f=4x%20%7B%7D%5E%7B2%7D%20%20%2B%2020x%20%2B%2025" id="TexFormula1" title="4x {}^{2} + 20x + 25

BigorU [14]

$\sf{\qquad\qquad\huge\underline{{\sf Answer}}}$

Let's Solve ~

$\qquad \sf \dashrightarrow \: 4 {x}^{2} + 20x + 25$

$\qquad \sf \dashrightarrow \: 4 {x}^{2} + 10x + 10x + 25$

$\qquad \sf \dashrightarrow \: 2x(2x + 5) + 5(2x + 5)$

$\qquad \sf \dashrightarrow \: (2x + 5) (2x + 5)$

$\qquad \sf \dashrightarrow \: (2x + 5) {}^{2}$

$\sf{\qquad \sf \dashrightarrow \: 4x² + 20x + 25 }$

$\sf{\qquad \sf \dashrightarrow \: (2x)² + (2 \sdot 2x \sdot 5) + (5)²}$

[ it's similar to expression - a² + 2ab + b² that is equal to (a + b)² ]

so, let's use this identity here to factorise :

$\sf{\qquad \sf \dashrightarrow \: (2x + 5)²}$

I hope it was helpful ~

3 0

1 year ago

Find the missing dimension of the triangle.

ludmilkaskok [199]

Answer: 16

Step-by-step explanation: 20/2 = 10

10x=160

10x/10=160/10

x=16

3 0

2 years ago

The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

$cos\theta_1$ is always positive in 4th quadrant and

$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

6 0

2 years ago

What operation should I use to solve -6x = -24. Addition, subtraction, multiplication,or division

olga_2 [115]

Always use PEMDAS as a key. Peretheses, Exponents, Multiplication, Division, Addition, and then Subtraction.

4 0

2 years ago

Suppose a triangle has sides a, b, and c and the angle opposite the side of length a is acute what must be true?

ANTONII [103]

The true statement about the triangle is (a) b^2 + c^2 > a^2

<h3>How to determine the true inequality?</h3>

The sides are given as:

a, b and c

The angle opposite of side length a is an acute angle

The above means that:

The side a is the longest side of the triangle.

The Pythagoras theorem states that:

a^2 = b^2 + c^2

Since the triangle is not a right triangle, and the angle opposite a is acute.

Then it means that the square of a is less than the sum of squares of other sides.

This gives

a^2 < b^2 + c^2

Rewrite as:

b^2 + c^2 > a^2

Hence, the true statement about the triangle is (a) b^2 + c^2 > a^2