Hi, prove that (AXB).(CXD)+(BXC).(AXD)+(CXA).(BXD)=0

△ABC has a>b>c. If one of the angles in the triangle is 60°, name the angle that can be equal to 90°.

Aleksandr-060686 [28]

Answer:

Angle A

Step-by-step explanation:

a is greater than b, which is greater than c

a triangle has 180 degrees

so 180-90 is 90

90-60 is 30

a>b>c

90>60>30

this would be a right triangle since one of the angles is 90 degrees

4 0

3 years ago

Read 2 more answers

. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0

3 years ago

Does anybody know the measure of JLM?

WARRIOR [948]

Measure of ∠JLM = m∠60°

6 0

2 years ago

The total income from the sale of cheeseburgers was what percent greater in 2004 than in 2003?

sveticcg [70]

$\bf \begin{array}{|c|c|c|ll} \cline{1-3} year&quantity&price\\ \cline{1-3} 2003&Q&P\\&&\\ 2004&20\%Q+Q&\\&&\\ &\frac{20}{100}Q+Q&\\&&\\ &1.2Q&\\&&\\ &&10\%P+P\\ &&\frac{10}{100}P+P\\&&\\ &&1.10P\\ \cline{1-3} &&\\ &(1.2Q)&(1.10P)\\ \cline{1-3} \end{array}\qquad \begin{array}{llll} \stackrel{\textit{revenue for 2003}}{(Q)(P)}\implies QP\\\\ \stackrel{\textit{revenue for 2004}}{(1.2Q)(1.10P)}\implies 1.32QP \end{array}$

7 0

4 years ago

3x+3y=10<br> -9x-9y=-30<br> please help

tiny-mole [99]

Answer:

Infinitely many solutions.

Explanation:

In this case we gonna use ELIMINATION BY ADDITION method.For that first we are gonna eliminate the terms containing ( X ).

Equation no 1:

3x + 3y = 10

Equation no 2:

-9x - 9y = -30

Now multiply equation no 1 with (3 )

3(3x + 3y) = 3(10)

9x + 9y = 30 ( equation no 1 )

Now ADD both the equations

9x + 9y = 30

<u> -9x - 9y =-30</u>

0 = 0

6 0

3 years ago