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4 years ago

What type of trig has the equation ax+bx+c=0 ?

Mathematics

1 answer:

Ber [7]4 years ago

4 0

Answer:

If it is ax^2 + bx + c = 0 then it is a quadratic equation

Step-by-step explanation:

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34c is otside in the sun and 12c is inside the frezzer so what is the differnce

-BARSIC- [3]

34-12=22
the answer is:
If the outside is 34c and the inside is 12c then the answer is 22c, as you are taking off 12 from 34.

8 0

3 years ago

X+y=5 and x-3y=3 linear equation I don't get an answer that makes sense

ddd [48]

If you want it to be y=mx+b form then the equations should look like this: y=x+5 and y=-1/3x+1.

3 0

3 years ago

Question: F=d+e+t, solve for t.

Komok [63]

F = d + e + t Flip the sides to put t on the left
d + e + t = F Subtract d from both sides
e + t = F - d Subtract e from both sides
t = F - d - e

7 0

3 years ago

Read 2 more answers

10. A 50-ft ladder leans against a building so that the base of the ladder is 15 feet from the

Hunter-Best [27]

By applying the trigonometry ratio, SOH, the angle that the ladder makes with the building is calculated as: 17.5°

<em><u>Recall:</u></em>

Trigonometry ratios that can be used to solve a right triangle are: SOH CAH TOA.

SOH represents: sin ∅ = Opp/Hyp

CAH represents: cos ∅ = Adj/Hyp

TOA represents: tan ∅ = Opp/Adj

The diagram attached below depicts the problem given.

∅ = x

Opp = 15 ft

Hyp = 50 ft

Thus, applying the trigonometry ratio, SOH, we have:

sin x = 15/50

x = $sin^{-1}(15/50)$

x = 17.5°

In conclusion, by applying the trigonometry ratio, SOH, the angle that the ladder makes with the building is calculated as: 17.5°

Learn more about trigonometry ratio on:

brainly.com/question/4326804

6 0

2 years ago

Alenkasestr [34]

We start with the expression at the left of the equation.

We can combine the terms as:

$\begin{gathered} \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}} \\ \frac{2+\sqrt[]{3}}{\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})}-\frac{2-\sqrt[]{3}}{\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}}}\cdot\frac{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})} \end{gathered}$

We can now apply the distributive property for the both the numerator and denominator. We can see also that the denominator is the expansion of the difference of squares:

$\begin{gathered} \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})-(2-\sqrt[]{3})\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{(\sqrt[]{2})^2-(\sqrt[]{2-\sqrt[]{3}}))^2} \\ \frac{(2+\sqrt[]{3})\cdot(\sqrt[]{2}-\sqrt[]{2-\sqrt[]{3}})+(\sqrt[]{3}-2)\cdot(\sqrt[]{2}+\sqrt[]{2-\sqrt[]{3}})}{2^{}-(2-\sqrt[]{3})^{}} \\ \frac{\sqrt[]{2}\cdot(2+\sqrt[]{3})-\sqrt[]{2-\sqrt[]{3}}\cdot(2+\sqrt[]{3})+\sqrt[]{2}\cdot(\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}\cdot(\sqrt[]{3}-2)}{2-2+\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2+\sqrt[]{3}+\sqrt[]{3}-2)+\sqrt[]{2-\sqrt[]{3}}(-2-\sqrt[]{3}+\sqrt[]{3}-2)}{\sqrt[]{3}} \\ \frac{\sqrt[]{2}(2\sqrt[]{3})+\sqrt[]{2-\sqrt[]{3}}(-4)}{\sqrt[]{3}} \\ 2\sqrt[]{2}-4\frac{\sqrt[]{2-\sqrt[]{3}}}{\sqrt[]{3}} \end{gathered}$

We then can continue rearranging this as:

7 0

1 year ago