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serg [7]
3 years ago
5

20r3 − 16r2 + 15r − 12

Mathematics
1 answer:
kati45 [8]3 years ago
7 0

Answer:

(5r-4)(4r^(2)+3)

Step-by-step explanation:

Hope it helps

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An isosceles triangle ABC has legs of length 24 and a vertex angle that measures 36º . Determine the length of its base, BC , to
iris [78.8K]

Answer: BC=14.8

Step-by-step explanation:

By definition, an Isosceles triangle has two equal sides and its opposite angles are congruent.

Observe the figure attached, where the isosceles triangle is divided into two equal right triangles.

So, in this case you need to use the following Trigonometric Identity:

sin\alpha =\frac{opposite}{hypotenuse}

In this case, you can identify that:

\alpha =\angle BAD=\angle CAD=18\°\\\\opposite=BD=CD=x\\\\hypotenuse=AB=AC=24

Substituting values, and solving for "x", you get:

sin(18\°)=\frac{x}{24}\\\\24*sin(18\°)=x\\\\x=7.416

Therefore, the length of BC rounded to nearest tenth, is:

BC=2(7.416)=14.8

6 0
3 years ago
Help pls!! math.. PLS HELPPP
AURORKA [14]

Answer:a

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Please help me out............................
Alina [70]

Answer:

<h2>b = 15°</h2>

Step-by-step explanation:

If Pq = RQ then ΔPQR is the isosceles triangle. The angles QPR and PRQ have the same measures.

We know: The sum of the measures of the angeles in the triangle is equal 180°. Therefore we have the equation:

m∠QPR + m∠PRQ + m∠RQP = 180°

We have

m∠QPR = m∠PRQ and m∠RQP = 60°

Therefore

2(m∠QPR) + 60° = 180°       <em>subtract 60° from both sides</em>

2(m∠QPR) = 120°           <em>divide both sides by 2</em>

m∠QPR = 60° and m∠PRQ = 60°

Therefore ΔPRQ is equaliteral.

ΔPSR is isosceles. Therefore ∠SPR and ∠PRS are congruent. Therefore

m∠SPR = m∠PRS

In ΔAPS we have:

m∠SPR + m∠PRS + m∠RSP = 180°

2(m∠SPR) + 90° = 180°            <em>subtract 90° from both sides</em>

2(m∠SPR) = 90°             <em>divide both sides by 2</em>

m∠SPR = 45° and m∠PRS = 45°

m∠PRQ = m∠PRS + b

Susbtitute:

60° = 45° + b           <em>subtract 45° from both sides</em>

15° = b

3 0
3 years ago
Caleb has an assignment to draw a contour map. What type of landform is he most likely drawing?
marysya [2.9K]

The type of land form that Caleb most likely draws is a hill. The answer is option C- a hill.

Step-by-step explanation:

  • Contour lines are <u>drawn to represent the changes in the elevation</u> which helps to know the slope of the figure drawn.
  • If the contour lines are drawn close, then steeper the figure is. And more distant the lines are drawn, then less steep the figure is.
  • Here, the lines seem to be likely circle in shape. The distance of the outermost contour line is 20 ft. The <u>next inner lines will have reduced values of 20 ft.</u>
  • Therefore, <u>the elevation gets reduced from higher to lower points</u> which provides a diagram of a hill. Since there is no continuous high and low elevations, this cannot be a valley (a range of mountains).
  • Furthermore, The contour diagram for option A and D such as creek(water stream) or shoreline will have different and mixed values of elevations that seems to be a messy structure of contour lines.
8 0
3 years ago
How do u do b??????????
horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

a)

if she uses and input of 6 on the 2nd one, the result will be 6² - 6  = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

b)

clearly we can never get an output  of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5

5 0
3 years ago
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