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lana [24]
2 years ago
12

⚠️‼️ HELP PLEASE GIVING POINTS ​

Mathematics
1 answer:
Nezavi [6.7K]2 years ago
8 0

Answer:

CBD = 63 degrees

Step-by-step explanation:

To find CBD, we must first find the value of x. To do so, we use this equation since CBD and ABC equal 96 degrees:

(6x+27) + (7x-9) = 96
13x+18 = 96
13x+18-18 = 96-18
13x = 78
13x/13 = 78/13
x = 6

Now that we know the value of x, we factor it in to CBD to find its true value:

6x+27
6(6)+27
36+27
= 63 degrees

To check if this answer is correct, we can also find the value of ABC and see if the two values equal 96:

7x-9
7(6)-9
42-9
= 33 degrees

63 + 33 = 96

The answer is correct. CBD = 63 degrees and ABC = 33 degrees.

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3.) A linear function passes through the points (0, 2) and (-4,0).
kotegsom [21]

Answer:

Step-by-step explanation:

The x-intercept is where y is 0.

The y-intercept is where x is 0.

The x-intercept would be (-4, 0) and the y-intercept (0, 2)

Using the slope formula, (y2-y1/x2-x1), we can find the slope which is 1/2.

7 0
3 years ago
The factors of m^2 + 12m + 35 are?
valina [46]
M^2 + 12m + 35

Factored:
(m+5)(m+7)

Check:
(m+5)(m+7)
(m*m) + (m*7) + (5*m) + (5*7)
m^2 + 12m + 35

Hope this helps! :)
5 0
3 years ago
PleAse help and show how you got it if you don’t mind
Lorico [155]
1)

\bf \textit{surface area of a cube}\\\\
s=6x^2\quad 
\begin{cases}
x=\textit{length of a side}\\
---------\\
s=150
\end{cases}\implies 150=6x^2\implies \cfrac{150}{6}=x^2
\\\\\\
25=x^2\implies \sqrt{25}=x\implies \boxed{5=x}

3)

\bf \textit{volume of a square pyramid}\\\\
V=\cfrac{b^2h}{3}\quad 
\begin{cases}
b=\textit{base's side}\\
h=height\\
-------\\
h=3\\
V=25
\end{cases}\implies 25=\cfrac{b^2\cdot 3}{3}\implies 25=b^2\cdot \cfrac{3}{3}
\\\\\\
25=b^2\cdot 1\implies 25=b^2\implies \sqrt{25}=b\implies \boxed{5=b}
3 0
3 years ago
Find all solutions in the interval from [0,2pi)<br> 2cos(3x)= -sqrt{2}
algol [13]

Solutions of 2cos(3x)= -\sqrt{2} in the interval from [0,2pi) is x =\frac{\pi}{12}  and x = \frac{23\pi}{12} .

<u>Step-by-step explanation:</u>

Find all solutions in the interval from [0,2pi)

2cos(3x)= -\sqrt{2}

⇒ 2cos(3x)= -\sqrt{2}

⇒ \frac{2cos(3x)}{2}= \frac{-\sqrt{2}}{2}

⇒ cos3x= \frac{-\sqrt{2}(\sqrt{2})}{2{\sqrt{2}}}

⇒ cos3x= \frac{-2}{2{\sqrt{2}}}

⇒ cos3x= \frac{-1}{{\sqrt{2}}}

⇒ cos^{-1}(cos3x)= cos^{-1}(\frac{-1}{{\sqrt{2}}})

⇒ 3x=\pm \frac{\pi}{4}

⇒ x=\pm \frac{\pi}{12}

Cosine General solution is :

x = \pm cos^{-1}(y)+ 2k\pi

⇒ x = \pm \frac{\pi}{12}+ 2k\pi , k is any integer .

At k=0,

⇒ x =\frac{\pi}{12} ,

At k=1,

⇒ x = - \frac{\pi}{12}+ 2\pi

⇒ x = \frac{23\pi}{12}

Therefore , Solutions of 2cos(3x)= -\sqrt{2} in the interval from [0,2pi) is x =\frac{\pi}{12}  and x = \frac{23\pi}{12} .

5 0
3 years ago
If f(x) = x2 - 4x and f(-3), then the result is:<br> o -3.<br> O 21.<br> 0 -21.
Vladimir79 [104]
This is pretty much saying that if x = -3, then solve for the equation, so (-3)^2 - 4(-3) = 9 - 12 = -3
3 0
3 years ago
Read 2 more answers
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