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3 years ago

For the given true statements, what can you conclude?

Mathematics

2 answers:

grandymaker [24]3 years ago

6 0

Answer:

D:QR+RS=QS

Step-by-step explanation:

We are given that

If point A, B and C are collinear, B is between A nd C, then AB+BC=AC

Points Q, R and S are collinear and R is between Q and S.

We have to find the conclusion about given statement.

If R lie between Q and R

Then, QR+RS=QS (segment addition property)

Because point Q, R and S are collinear.

Therefore, option D is true.

Answer:D:QR+RS=QS

sergey [27]3 years ago

4 0

The answer is QR+RS=QS

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Simora [160]

The standard form of a quadratic equation is ,

ax² + bx + c = 0.

And the formula to find the discriminant is b² - 4ac.

Here the first step is to change the given equation into standard form. So, add 1 to each sides of the equation. Therefore,

2x² – 9x + 2+1 = –1 + 1

2x² – 9x + 3 = 0

Next step is to compare the given equation with this equation to get the value of a, b and c.

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3 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

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A source of laser light sends rays AB and AC toward two opposite walls of a hall. The light rays strike the walls at points B an

Nataly [62]

Answer:

The distance between the walls is 70 m.

Step-by-step explanation:

Given: A source of laser light is at point A on the ground between two parallel walls BE and CD . The walls are perpendicular to the ground that is

BE ⊥ ED and CD ⊥ ED

AB is a ray of light which strikes the wall on the left at point B which is 30 meters above the ground. that is BE = 30 m

AC is a ray of light which strikes the wall on the right at point C. The length of AC = 80 meters.

The ray AB makes an angle of 45 degrees with the ground that is m∠BAE = 45°

The ray AC makes an angle of 60 degrees with the ground that is m∠CAD = 60°

As shown is figure attached below.

WE have to find the distance between the walls that is Length of ED

Length of ED = EA + AD

Consider the Δ AEB,

Using trigonometric ratio,

$\tan\theta=\frac{perpendicular}{base}$

Here $\theta=45^{\circ}$ , perpendicular = 30 m and base we can find.

thus,

$\tan 45^{\circ}=\frac{30}{EA}$

We know $\tan 45^{\circ}=1$

thus, EA = 30 m

Consider the Δ AEB,

Using trigonometric ratio,

$\cos\theta=\frac{base}{hypotenuse}$

Here $\theta=60^{\circ}$ , hypotenuse = 80 m and base we can find.

thus, $\cos 60^{\circ}=\frac{base}{80}$

We know, $\cos 60^{\circ}=\frac{1}{2}$

thus, Base = 40 m

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3 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

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3 years ago

Algebraically solve the inequality: 5x-6>/1+2(4x+1)

elena-14-01-66 [18.8K]

Answer:

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Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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3 years ago