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Alex17521 [72]
3 years ago
11

(PLZ ASAP ANSWER !!!!!!)The number line shows the graph of an inequality: A number line is shown from negative 5 to positive 5 w

ith increments of 0.5. All the whole numbers are labeled on the number line. An empty circle is shown on the first mark to the left of 0. The region to the left of the empty circle is shaded. Which statement explains whether −2.5 can be a value in the shaded region? (5 points) Yes it can, because −2.5 lies to the left of −0.5. Yes it can, because −2.5 lies to the right of −0.5. No it cannot, because −2.5 lies to the left of −0.5. No it cannot, because −2.5 lies to the right of −0.5.
Mathematics
1 answer:
goblinko [34]3 years ago
3 0

it is a. Yes it can, because −2.5 lies to the left of −0.5. plz mark me as brainiest and tell me if I am wrong

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Decimals that have the same value are
shepuryov [24]
<span>Decimals that have the same value are equivalent decimals. Equivalent decimals are decimal numbers that have the same value (the same amount). In other words equivalent decimal numbers have the same value but different number of decimals. For example: 
0.5 and 0.50 are equivalent decimals.
0.5 and 0.500
0.05 and 0.0500

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4 0
3 years ago
A balloon is rising at a constant speed of 5 ftys. A boy is cycling along a straight road at a speed of 15 ftys. When he passes
Ket [755]

Answer:

  13 ft/s

Step-by-step explanation:

t seconds after the boy passes under the balloon the distance between them is ...

  d = √((15t)² +(45+5t)²) = √(250t² +450t +2025)

The rate of change of d with respect to t is ...

  dd/dt = (500t +450)/(2√(250t² +450t +2025)) = (50t +45)/√(10t² +18t +81)

At t=3, this derivative evaluates to ...

  dd/dt = (50·3 +45)/√(90+54+81) = 195/15 = 13

The distance between the boy and the balloon is increasing at the rate of 13 ft per second.

_____

The boy is moving horizontally at 15 ft/s, so his position relative to the spot under the balloon is 15t feet after t seconds.

The balloon starts at 45 feet above the boy and is moving upward at 5 ft/s, so its vertical distance from the spot under the balloon is 45+5t feet after t seconds.

The straight-line distance between the boy and the balloon is found as the hypotenuse of a right triangle with legs 15t and (45+5t). Using the Pythagorean theorem, that distance is ...

  d = √((15t)² + (45+5t)²)

7 0
3 years ago
Lcm of 2 and 3 very very very urgent
Licemer1 [7]

Answer: 6

Step-by-step explanation: 2x3=6 it’s the lowest multiple

5 0
2 years ago
Read 2 more answers
Simplify:<br> 2 11/12 + 5 1/2 + 3 3/8<br> Write you answer as a mixed number.
zysi [14]

Answer:

2\frac{11}{12} +5 \frac{1}{2} + 3\frac{3}{8} (write them out and convert them all into mixed fractions)

\frac{35}{12} + \frac{11}{2} + \frac{27}{8} (find the LCD aka least common denominator)

\frac{70}{24} + \frac{132}{24}  + \frac{81}{24} (add all numerators together)

\frac{283}{24}  (this is improper, so you need to convert back to mixed by dividing)

11\frac{19}{24} (this is your answer)

3 0
3 years ago
Suppose ABC is a right triangle with sides​ a, b, and c and right angle at C. Find the unknown side length using the Pythagorean
Leno4ka [110]

Answer:

a = \sqrt{7}

Step-by-step explanation:

I think your question is missed of key information, allow me to add in and hope it will fit the original one.:

<em>Suppose ABC is a right triangle with sides​ a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the values of the six trigonometric functions for angle B. when b=3 and c=4</em>.

My answer:

We will Pythagoras theorem, which states that the sum of squares of two legs of a right triangle is equal to the square of the hypotenuse of right triangle. Because the question says that ABC is a right triangle.

a^2+b^2=c^2

Given that: b=3 and c=4

a^2+3^2=4^2a^2+9=16a^2+9-9=16-9a^2=7

so a = \sqrt{7}

We know that tangent relates opposite side of a right triangle with adjacent side.

\text{tan}(B)=\frac{a}{b}\\\text{tan}(B)=\frac{\sqrt{7}}{3}

Please have a look at the attached photos.

3 0
3 years ago
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