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jonny [76]
3 years ago
5

If ∧ ASC ≅ ∧ KSC then ∠8 is congruent to which angle?

Mathematics
1 answer:
zalisa [80]3 years ago
7 0

Answer:

B. angle 5

Step-by-step explanation:

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First term is 3, common difference is 4 what is the 5th term
patriot [66]

Answer:

19

Step-by-step explanation:

The n th term of an arithmetic sequence is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 3 and d = 4 , thus

a₅ = 3 + (4 × 4) = 3 + 16 = 19

5 0
3 years ago
What two numbers that if multiplied equal -90 and if added equal 8?
Galina-37 [17]
x-first\ number\\y-second\ number\\\\ \left \{ {{x\cdot\ y=-90} \atop {x+y=8}} \right.\\\\From\ second\ equayion:\\x=8-y\\\\Substracted\ to\ first\ equation:\\(8-y)y=-90\\-y^2+8y+90=0\\\\\Delta=b^2-4ac\\\Delta=8^2-4\cdot(-1)\cdot90=64+360=424\\\sqrt{\Delta}=\sqrt{424}2\sqrt{106}\\\\y_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-8+2\sqrt{106}}{2\cdot(-1)}=4-\sqrt{106}\\y_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-8-2\sqrt{106}}{2\cdot(-1)}=4+\sqrt{106}\\\\x_1=8-(4-\sqrt{106})=8-4+\sqrt{106}=4+\sqrt{106}
x_2=8-(4-+\sqrt{106})=8-4-\sqrt{106}=4-\sqrt{106}\\\\ \left \{ {{y=4-\sqrt{106}} \atop {x=4+\sqrt{106}}} \right.\\

3 0
3 years ago
Help please! This is due in 3 minutes!
REY [17]
It’s 26. Hope that helped.
6 0
3 years ago
Read 2 more answers
What is an equation that shows that two ratios are equivalent?
Debora [2.8K]
A equation that shows two ratios are equivalent is a " true proportion"
3 0
3 years ago
Show that a vector
Shtirlitz [24]

Answer:

 u = |u|(cos∝+cosβ+cosγ)

Step-by-step explanation:

<u>Explanation</u>

<u>Proof:-</u>

Given a vector  u = x₁ i + y₁j +z₁k

let O X, OY, O Z be the positive co-ordinate axes

P(x₁,y₁,z₁) be any point in the space

Let OP makes angles α,β,γ with co-ordinate axes OX , OY ,OZ .

The angle α,β,γ are known as direction angles and cosine of the angle

l =cosα , m= cosβ , n=cosγ

The perpendicular PA,PB,PC are drawn co-ordinate axes OX,OY,OZ respecctively

InΔOAP , ∠A =90° , cos∝ =\frac{x}{r}

                                 x₁ = rcos∝

InΔOBP , ∠B =90° , cosβ =\frac{y}{r}

                                 y₁ = rcosβ

InΔOCP , ∠C =90° , cosγ =\frac{z}{r}

                                z₁ = rcosγ

Given   u = x₁ i + y₁j +z₁k

      |u| = \sqrt{(x_{1})^{2} +(x_{2} )^{2} +(x_{3} )^{2}  }

Therefore  u = x₁ i + y₁j +z₁k

               u = |u|(cos∝+cosβ+cosγ)

         

​

6 0
3 years ago
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