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zheka24 [161]
3 years ago
7

Plz anserw the question below.

Mathematics
1 answer:
yawa3891 [41]3 years ago
7 0

Answer:

x=70

Step-by-step explanation:

Due to the law of alternate interior angles, we can see that one of the angles in the triangle is equal to 35 degrees as well (please see the picture attached).

We're given that one of the exterior angles of the triangle is equal to 105. The exterior angle of a triangle will always be equal to the sum of the two opposite interior angles. Knowing this, we can write:

x+35=105

Subtract both sides by 35

x+35-35=105-35

x=70

Therefore, the value of x is 70 degrees.

I hope this helps!

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What is the ratio for cos B
nataly862011 [7]

Answer:

cos B = 8/17

Step-by-step explanation:

The cosine function is defined as (adjacent side) / (hypotenuse), and in this case is 8/17.

6 0
3 years ago
Please Help !!!
solong [7]
Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)

moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g


Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05

N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
5 0
4 years ago
Can anyone help with this?!?!
Gnom [1K]
Since y=5 is an asymptote, it means that y will approach 5 but never be 5. So you use (. Then, you see the function increasing from the asymptote so it would be (5, infinity) Choice C
5 0
3 years ago
Log(x⁴+3x³) - log(X + 3 ) + log2 - log6 = 2logx . find the value of x
AfilCa [17]

The given equation is

\begin{gathered} \log (x^4+3x^3)-\log (x+3)+\log 2-\log 6=2\log x \\ \log (\frac{x^4+3x^{3^{}}}{x+3})+\log \frac{2}{6}=\log x^2 \\ \log \frac{x^3(x+3)}{x+3}+\log \frac{1}{3}=\log x^2 \\ \log x^3+\log \frac{1}{3}=\log x^2 \\ \log \frac{x^3}{3}=\log x^2 \\ \frac{x^3}{3}=x^2 \\ x^3-3x^2=0 \\ x^2(x-3)=0 \end{gathered}

hence

x=0\text{ or x=3}

But x cannot be zero so x=3

So the value of x is 3

h

7 0
1 year ago
Please help<br> if x = 3 and y=-4 find x2 + y2 and x2 -y2
lions [1.4K]
1. Plug in the 3 in all the x’s that are in the equation:

(3)2 + y2 = for x2 + y2

(3)2 - y2 = for x2 - y2

This would get you for both equations..

6 + y2 = for x2 + y2

6 - y2 = for x2 - y2

2. Now let’s plug in the value of y which is -4:

6 + (-4)2 = for x2 + y2

6 - (-4)2 = for x2 + y2

This would get you for both equations...

-2 = x2 + y2

14 = x2 - y2


Hope this helped, have an awesome day!





6 0
3 years ago
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