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2 years ago

Which of the following notations correctly describe the end behavior of the polynomial graphed below

Mathematics

2 answers:

Naddika [18.5K]2 years ago

7 0

Answer:

B is the answer

Step-by-step explanation:

Irina18 [472]2 years ago

5 0

The answer is B
To visualize this, think of the first line as saying "As x approaches negative infinity, f(x) (or just y) approaches __". As the graph goes left, it goes up: so, as x approaches negative infinity, f(x) approaches infinity. As the graph goes right, it goes down: so, as x approaches infinity, f(x) approaches negative infinity.

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Which descriptions from the list below accurately describe the relationship between QRS and TUV? Check all that apply.

valentina_108 [34]

Definition: Triangles are congruent when they have exactly the same three sides and exactly the same three angles.

From the diagram you have that:

$m\angle Q=m\angle T=50^{\circ}$
$m\angle R=m\angle U=83^{\circ}$
$m\angle S=m\angle V=47^{\circ}$
$QR=TU=22$
$RS=UV=23$
$QS=TV=30$

Triangles QRS and TUV have congruent corresponding sides and congruent corresponding angles. This means they are congruent triangles. Congruent triangles always have the same shape and the same size. Also each pair of congruent triangles is a pair of similar triangles with ratio of similarity equal to 1.

Answer: all choices are correct.

4 0

1 year ago

Read 2 more answers

The hypotenuse of a 45°-45°-90° triangle measures 10 StartRoot 5 EndRoot in. A right triangle is shown. The hypotenuse has a len

suter [353]

Answer:

$5\sqrt{10}\ units$

5 StartRoot 10 EndRoot

Step-by-step explanation:

we know that

The legs of a 45°-45°-90° triangle are congruent

Let

x ----> the length of one leg of the triangle

Applying the Pythagorean Theorem

$c^2=a^2+b^2$

where

c is the hypotenuse

a and b are the legs

we have

$c=10\sqrt{5}\ units$

$a=b=x\ units$

substitute

$(10\sqrt{5})^2=x^2+x^2$

$500=2x^2\\x^2=250\\x=\sqrt{250}\ units$

Simplify

$x=5\sqrt{10}\ units$

5 StartRoot 10 EndRoot

4 0

2 years ago

Read 2 more answers

Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu

NeX [460]

Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

Let speed in clear weather = x

⇒ Speed in thunderstorm = x-20

Total time taken for trip = 1.5 hours

We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

Total time taken for trip = Time taken in clear weather + Time taken in thunderstorm

⇒ Total time taken for trip = $\frac{Distance covered in clear weather}{Speed in clear weather}$ + $\frac{Distance covered in thunderstorm}{Speed in thunderstorm}$

⇒ 1.5 = $\frac{50}{x}$ + $\frac{15}{x-20}$

⇒ 1.5 = $\frac{50(x-20)+15(x)}{(x)(x-20)}$

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = $\frac{-b+\sqrt{D}}{2a}$ ; x₂ = $\frac{-b-\sqrt{D}}{2a}$

⇒ x₁ = $\frac{-(-190)+\sqrt{12,100}}{2*3}$ and x₂ = $\frac{-(-190)-\sqrt{12,100}}{2*3}$

⇒ x₁ = $\frac{190+110}{6}$ and x₂ = $\frac{190-110}{6}$

⇒ x₁ = $\frac{300}{6}$ and x₂ = $\frac{80}{6}$

⇒ x₁ = 50 and x₂ = 13.33

So speed in clear weather can be 50 mph or 13.33 mph. However, we know that in thunderstorm was 20 mph less than speed in clear weather.

If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

7 0

2 years ago

A television camera at ground level is filming the lift-off of a space shuttle at a point 750 meters from the launch pad. Let θ

ss7ja [257]

750 * x =

0^1 = 0 =

750 x 0 = 750

A: 750 meters

7 0

1 year ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago