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3 years ago

-5=x/3-8 plz someone help me plz

1 answer:

5 0

Isolate x by adding 8 to both sides.
3 = x/3
Then, multiply by 3 on both sides.
9 = x
To prove that this is right, you can imput 9 into the equation and it equals -5.

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Is it true or false?

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False

Step-by-step explanation:

I'm pretty sure it's false, sry if it's not

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What decimal number is represented by the shaded part

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The decimal that represents the shaded portion of the figure is 0.75.

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2 years ago

Help me out please !!!!!

evablogger [386]

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Need help solving algebraically!

alekssr [168]

We need to account for both x values on either side of the length, and width.
Thus, the length becomes 10 + x + x = 10 + 2x
and the width becomes 5 + x + x = 5 + 2x

For the second question, I'm assuming we don't account for the area that is covered by the garden.
Then we can say that the path is measured by: (5 + 2x)(10 + 2x) - 50, which is the area of the garden itself.

(5 + 2x)(10 + 2x) - 50 = 54

Expanding the brackets:
$50 + 10x + 20x + 4x^{2} - 50 = 54$
$4x^{2} + 30x = 54$
$2x^{2} + 15x - 27 = 0$
$2x^{2} - 3x + 18x - 27 = 0$
$x(2x - 3) + 9(2x - 3) = 0$

$(x + 9)(2x - 3) = 0$
x = -9, or x = 3/2

Since x > 0, then x ≠ -9
Thus, the only x-value we can take is x = 3/2

6 0

3 years ago

This is a geometry question, i need something quickly :)

Marysya12 [62]

Answer:

hope it helps mark me brainlieast!

Step-by-step explanation:

For triangle ABC with sides a,b,c labeled in the usual way,

c2=a2+b2−2abcosC

We can easily solve for angle C .

2abcosC=a2+b2−c2

cosC=a2+b2−c22ab

C=arccosa2+b2−c22ab

That’s the formula for getting the angle of a triangle from its sides.

The Law of Cosines has no exceptions and ambiguities, unlike many other trig formulas. Each possible value for a cosine maps uniquely to a triangle angle, and vice versa, a true bijection between cosines and triangle angles. Increasing cosines corresponds to smaller angles.

−1≤cosC≤1

0∘≤C≤180∘

We needed to include the degenerate triangle angles, 0∘ and 180∘, among the triangle angles to capture the full range of the cosine. Degenerate triangles aren’t triangles, but they do correspond to a valid configuration of three points, namely three collinear points.

The Law of Cosines, together with sin2θ+cos2θ=1 , is all we need to derive most of trigonometry. C=90∘ gives the Pythagorean Theorem; C=0 and C=180∘ give the foundational but often unnamed Segment Addition Theorem, and the Law of Sines is in there as well, which I’ll leave for you to find, just a few steps from cosC= … above. (Hint: the Law of Cosines applies to all three angles in a triangle.)

The Triangle Angle Sum Theorem, A+B+C=180∘ , is a bit hard to tease out. Substituting the Law of Sines into the Law of Cosines we get the very cool

2sinAsinBcosC=sin2A+sin2B−sin2C

Showing that’s the same as A+B+C=180∘ is a challenge I’ll leave for you.

In Rational Trigonometry instead of angle we use spreads, squared sines, and the squared form of the formula we just found is the Triple Spread Formula,

4sin2Asin2B(1−sin2C)=(sin2A+sin2B−sin2C)2

true precisely when ±A±B±C=180∘k , integer k, for some k and combination of signs.

This is written in RT in an inverted notation, for triangle abc with vertices little a,b,c which we conflate with spreads a,b,c,

(a+b−c)2=4ab(1−c)

Very tidy. It’s an often challenging third degree equation to find the spreads corresponding to angles that add to 180∘ or zero, but it’s a whole lot cleaner than the trip through the transcendental tunnel and back, which almost inevitably forces approximation.

6 0

2 years ago