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3 years ago

Solve the system by graphing y = -3/4x + 6 y = 2x - 5

Mathematics

1 answer:

jonny [76]3 years ago

4 0

Answer:

X= 4. Y = 3.

Step-by-step explanation:

If you have a graphing calculator, put the first equation into Y1. And second equation into Y2. Then, find the intersection of the two equations. That's how you find your answer.

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In △ABC, D is a point on side AC¯¯¯¯¯¯¯¯ such that BD=DC and ∠BCD measures 70∘. What is the degree measure of ∠ADB?

777dan777 [17]

If angle BCD measures 70° then so does angle DBC (because you have formed an isoceles triangle inside the larger ΔABC and the two legs are equal so the two angles have to be equal. So we have a two 70° angles which leaves 40° for the 3rd angle, which is ∠BDC.

Since ∠BDC and ∠ADB are supplementary (180°) - that leaves 140° for ∠ADB and is our answer

5 0

3 years ago

My 3-digit has a 4 in the hundred place. It has a greater digit in the tens place than in the ones place. The sum of the digit i

Vika [28.1K]

420 hope that helped!!

3 0

3 years ago

Given that <img src="https://tex.z-dn.net/?f=I_0%20%3D%2010%5E-12" id="TexFormula1" title="I_0 = 10^-12" alt="I_0 = 10^-12" alig

Alex Ar [27]

The intensity of sound which the decibel level of the sound measures 156 is 3.981 × 10³ W/m²

Decibel level, dB = 10log₁₀(I/I₀) where dB = decibel level = 156, I = intensity at 156 dB and I₀ = 10⁻¹² W/m².

Since we require I, making I subject of the formula, we have

dB/10 = log₁₀(I/I₀)

$\frac{I}{I_{0} } = 10^{\frac{dB}{10} } \\I = I_{0} 10^{\frac{dB}{10} }$

Substituting the values of the variables into the equation, we have

$I = I_{0} 10^{\frac{dB}{10} }\\I = 10^{-12} 10^{\frac{156}{10} }\\I = 10^{-12} 10^{15.6}\\I = 10^{15.6-12} \\I = 10^{3.6} W/m^{2}$

I = 3981.072W/m²

I = 3.981 × 10³ W/m²

So, the intensity of sound which the decibel level of the sound measures 156 is 3.981 × 10³ W/m²

Learn more about intensity of sound here:

brainly.com/question/4431819

3 0

3 years ago

SIMPLIFY 1/4 × 1/3 - 1/5× 1/4 + 1/4 × 1/6

MrMuchimi

Answer:

3/40

Step-by-step explanation:

1/4 x 1/3 - 1/5 x 1/4 +1/4 x 1/6

(1/4 x 1/3) - (1/5 x 1/4) + (1/4 x 1/6)

1/12 - 1/20 + 1/24

1/30 + 1/24

3/40

6 0

3 years ago

Read 2 more answers

Prove that if n is a perfect square then n + 2 is not a perfect square

notka56 [123]

Answer:

This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$ .)

$\text{Let $n \in \mathbb{N}$ be a perfect square}$ .

$\textbf{Case 1.} ~ \text{n = 0}$ :

$\text{$n + 2 = 2$, which isn't a perfect square}$ .

$\text{Claim verified for $n = 0$}$ .

$\textbf{Case 2.} ~ \text{$n \in \mathbb{N}$ and $n \ne 0$. Hence $n \ge 1$}$ .

$\text{Assume that $n$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}$ .

$\text{Assume $\textit{by contradiction}$ that $(n + 2)$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}$ .

$\text{$n + 2 > n > 0$ $\implies$ $b = \sqrt{n + 2} > \sqrt{n} = a$}$ .

$\text{$a,\, b \in \mathbb{N} \subset \mathbb{Z}$ $\implies b - a = b + (- a) \in \mathbb{Z}$}$ .

$\text{$b > a \implies b - a > 0$. Therefore, $b - a \ge 1$}$ .

$\text{$\implies b \ge a + 1$}$ .

$\text{$\implies n+ 2 = b^2 \ge (a + 1)^2= a^2 + 2\, a + 1 = n + 2\, a + 1$}$ .

$\text{$\iff 1 \ge 2\,a $}$ .

$\text{$\displaystyle \iff a \le \frac{1}{2}$}$ .

$\text{Contradiction (with the assumption that $a \ge 1$)}$ .

$\text{Hence the original claim is verified for $n \in \mathbb{N}\backslash\{0\}$}$ .

$\text{Hence the claim is true for all $n \in \mathbb{N}$}$ .

Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ( $a \in \mathbb{N}$ ) such that $a^2 = n$ .

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$ .

Note, that since $(n + 2) > n \ge 0$ , $\sqrt{n + 2} > \sqrt{n}$ . Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$ , one can conclude that $b > a$ .

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$ . In other words, if $b > a$ , then it must be true that $b \ge a + 1$ .

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$ .)

$b^2 \ge (a + 1)^2$ .

Expand the right-hand side using the binomial theorem:

$(a + 1)^2 = a^2 + 2\,a + 1$ .

$b^2 \ge a^2 + 2\,a + 1$ .

However, recall that it was assumed that $a^2 = n$ and $b^2 = n + 2$ . Therefore,

$\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1$ .

$n + 2 \ge n + 2\, a + 1$ .

Subtract $n + 1$ from both sides of the inequality:

$1 \ge 2\, a$ .

$\displaystyle a \le \frac{1}{2} = 0.5$ .

Recall that $a$ was assumed to be a natural number. In other words, $a \ge 0$ and $a$ must be an integer. Hence, the only possible value of $a$ would be $0$ .

Since $a$ could be equal $0$ , there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that $a = 0$ just won't work as in the assumption.

If indeed $a = 0$ , then $n = a^2 = 0$ . $n + 2 = 2$ , which isn't a perfect square. That contradicts the assumption that if $n = 0$ is a perfect square, $n + 2 = 2$ would be a perfect square. Hence, by contradiction, one can conclude that

$\text{if $n$ is a perfect square, then $n + 2$ is not a perfect square.}$ .

Note that to produce a more well-rounded proof, it would likely be helpful to go back to the beginning of the proof, and show that $n \ne 0$ . Then one can assume without loss of generality that $n \ne 0$ . In that case, the fact that $\displaystyle a \le \frac{1}{2}$ is good enough to count as a contradiction.

7 0

3 years ago