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pishuonlain [190]
2 years ago
12

Which is the correct solution to ‐3x<21?

Mathematics
2 answers:
elena-s [515]2 years ago
8 0
Top right
this is because, when dividing an inequality by a negative number, you must flip the inequality sign

-3x < 21
x > -7
aleksandrvk [35]2 years ago
3 0

Answer:

Top right

Step-by-step explanation:

When mult/dividing by a negative number, the sign always flips.

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AC = <br> Round your answer to the nearest hundredth.
dusya [7]

Answer:

AC ≈ 5.03

Step-by-step explanation:

We can solve the problem above using the trigonometric ratio, they are;

SOH CAH TOA

sin Ф = opposite / hypotenuse

cosФ= adjacent/ hypotenuse

tan Ф = opposite / adjacent

From the diagram above, in reference to angle B;

opposite =AC     and   adjacent =BC

Since we have opposite and adjacent, the best formula to use is

tanФ = opposite / adjacent

tan B = AC  / BC

tan 40 =  AC/ 6

Multiply both-side of the equation by 6

6× tan 40   =  AC/ 6  ×  6

At the right-hand side of the equation, 6 will cancel-out 6 leaving us with just AC

6×tan 40 = AC

5.034598  = AC

AC ≈ 5.03   to the nearest hundredths

4 0
3 years ago
Which angle is complimentary to 3?
MA_775_DIABLO [31]

Answer:

non of yhrm

Step-by-step explanation:

8 0
2 years ago
The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0
2 years ago
What is the fraction of 10-6
gladu [14]

Answer:

1 4/6

Step-by-step explanation:

8 0
3 years ago
A 5 metre long ladder is placed against a wall. It reaches
Juliette [100K]
<h3><em><u>given</u></em></h3>

<em><u>5m</u></em><em><u> </u></em><em><u>long</u></em><em><u> </u></em><em><u>ladder</u></em><em><u> </u></em><em><u>placed</u></em><em><u> </u></em><em><u>against</u></em><em><u> </u></em><em><u>4.3m</u></em><em><u> </u></em><em><u>up</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>wall</u></em><em><u>.</u></em>

<h3><em><u>to</u></em><em><u> </u></em><em><u>find</u></em><em><u>:</u></em></h3>

<em><u>the</u></em><em><u> </u></em><em><u>angle</u></em><em><u> </u></em><em><u>between</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>ladder</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>ground</u></em><em><u>.</u></em>

<h3><em><u>solution</u></em><em><u>:</u></em></h3>

<em><u>Based on the given conditions, formulate: </u></em>

<em><u>sin⁡x=4.3÷5</u></em>

<em><u>Evaluate the equation/expression</u></em><em><u>:</u></em>

<em><u>x</u></em><em><u>=</u></em><em><u> </u></em><em><u>1.0352</u></em><em><u>7</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>2.10</u></em><em><u>6</u></em><em><u>3</u></em><em><u>2</u></em>

<em><u>x= 1.03527</u></em><em><u>°</u></em>

<em><u>or</u></em><em><u> </u></em><em><u>x</u></em><em><u>=</u></em><em><u> </u></em><em><u>2.10632</u></em><em><u>°</u></em>

6 0
2 years ago
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