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2 years ago

If p is true and q is false, the p ---> q is (always, never, sometimes) true.

Mathematics

2 answers:

garri49 [273]2 years ago

5 0

I believe the answer your look for is ....[ALWAYS]

natulia [17]2 years ago

5 0

Answer:

The correct answers are filled up below:

Step-by-step explanation:

If p is true and q is false, the p ---> q is (always) true.

When p is false and q is true, then p or q is (always) true.

If p is true and ~ q is false, then p ---> ~ q is (always, ) false.

If p is true and q is true, then ~ p ---> ~ q is (always) true.

If p ---> q is true and q is true, then p is (always, ) true

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Which model represents the expression

NARA [144]

Answer:

Step-by-step explanation:

Because you add a denominator of 1 to the whole number and just multiply over and you get 6/8

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2 years ago

Solve for x. x-17=-57<br> i need help

seraphim [82]

X - 17 = -57
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3 years ago

The box and whiskers plot represents the results of the same math test being given to two different classes. According to the bo

densk [106]

Answer: Answer A: Both classes have a range of 45

Step-by-step explanation:

above the two graphs we can see a number line ranging from 50-100

bellow that, we can see 2 box and whisker plots, one for second period and one for fourth period. the larger rectangular boxes show all the number results, the line in between them states the average result. Both lines have starting points at 55 and ending points at 100.

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B is wrong because 2nd period has a median of 90

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D is wrong because most of the students scored an 85+

conclusion: answer A is correct

4 0

3 years ago

Read 2 more answers

Use the discriminant to determine what type of roots the equations will have, and categorize the equations according to their ro

topjm [15]

Step-by-step explanation:

The discriminant of the quadratic equation $ax^2+bx+c=0$ :

$\Delta=b^2-4ac$

If Δ < 0, then the equation has two complex roots $x=\dfrac{-b\pm\sqrt\Delta}{2a}$

If Δ = 0, then the equation has one repeated root $x=\dfrac{-b}{2a}[/tex If Δ > 0, then the equation has two discint roots [tex]x=\dfrac{-b\pm\sqrt\Delta}{2a}$

$1.\ x^2-4x+2=0\\\\a=1,\ b=-4,\ c=2\\\\\Delta=(-4)^2-4(1)(2)=16-8=8>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\\\\x=\dfrac{-(-4)\pm2\sqrt2}{2(1)}=\dfrac{4\pm2\sqrt2}{2}=2\pm\sqrt2\\\\==============================\\\\2.\ 5x^2-2x+3=0\\\\a=5,\ b=-2,\ c=3\\\\\Delta=(-2)^2-4(5)(3)=4-60=-56$

$3.\ 2x^2+x-6=0\\\\a=2,\ b=1,\ c=-6\\\\\Delta=1^2-4(2)(-6)=1+48=49>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt{49}=7\\\\x=\dfrac{-1\pm7}{(2)(2)}\\\\x_1=\dfrac{-8}{4}=-2,\ x_2=\dfrac{6}{4}=\dfrac{3}{2}\\\\==============================\\\\4.\ 13x^2-4=0\qquad\text{add 4 to both sides}\\\\13x^2=4\qquad\text{divide both sides by 13}\\\\x^2=\dfrac{4}{13}\to x=\pm\sqrt{\dfrac{4}{13}},\ \bold{two\ distinct\ roots}\\\\==============================$

$5.\ x^2-6x+16=0\\\\a=1,\ b=-6,\ c=16\\\\\Delta=(-6)^2-4(1)(16)=36-64=-28$

$7.\ 4x^2+11=0\qquad\text{subtract 11 from both sides}\\\\4x^2=-11\qquad\text{divide both sides by 4}\\\\x^2=-\dfrac{11}{4}\to x=\pm\sqrt{-\dfrac{11}{4}}\\\\x=\pm\dfrac{\sqrt{11}}{2}\ i,\ \bold{two\ complex\ roots}$

6 0

3 years ago

Read 2 more answers

Someone help please and please make sure it’s right :)

Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

5x + 5 = 4x + 16 ( being alternate angles)

5x - 4x = 16 - 5

x = 11

Hope it will help :)

6 0

2 years ago