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3 years ago

Which of the following illustrates the truth value of the following conditional?

Mathematics

2 answers:

Ilya [14]3 years ago

7 0

The answer to this would be FF -> T

ryzh [129]3 years ago

5 0

Answer:

F F → T

Step-by-step explanation:

Since, the Boolean truth table is,

p q p → q

T T T

T F F

F T F

F F T

Since, p: 7 + 1 = 0

⇒ p is false

q: 2 + 2 = 5

⇒ q is false,

Thus, the truth value of the given conditional is,

F F → T

First option is correct.

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What is the slope-intercept form of the linear equation 4x + 2y = 24? Drag and drop the appropriate number, symbol, or variable

Morgarella [4.7K]

The slope-intercept form of the linear equation $4x + 2y = 24$ is $y =-2x+12$

Step-by-step explanation:

We need to write slope-intercept form of the linear equation $4x + 2y = 24$

The standard equation of slope-intercept form is:

$y=mx+b$

Converting given equation in slope-intercept form.

$4x + 2y = 24$

Add -4x on both sides

$2y = -4x+24$

Divide both sides by 2

$y =-2x+12$

So, the slope-intercept form of the linear equation $4x + 2y = 24$ is $y =-2x+12$

Keywords: Slope-intercept form

Learn more about Slope-intercept form at:

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6 0

3 years ago

Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question.

makkiz [27]

Let

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y------> the number of free response question

we know that

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Substitute equation B in equation A

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3 0

1 year ago

Solve h^2=100 by the square root property if possible

hram777 [196]

H^2=100

H=10

(Because 10*10=100)

3 0

3 years ago

An airplane flies due north from ikeja airport for 500km.It then flies on a bearing of 060 from a further distance of 300km befo

Nata [24]

Answer:

482 km

63.94 degrees

Step-by-step explanation:

to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.

Cosine rule states that

C = $\sqrt{A^{2} + B^{2} -2AB cos(c) }$

where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060

solving now, we have

C = $\sqrt{500^{2} + 300^{2} -2 * 500 * 300 cos(60) }$

C = $\sqrt{250000 + 90000 - [215000 cos(60) }]$

C = $\sqrt{340000 - [215000 * 0.5 }]$

C = $\sqrt{340000 - [107500 }]$

C = $\sqrt{232500}$

C = 482 km

The bearing can be gotten by using the Sine Rule.

$\frac{sina}{A}$ = $\frac{sinc}{C}$

sina/500 = sin60/482

482 sina = 500 sin60

sina = $\frac{500 sin60}{482}$

sina = 0.8983

a = sin^-1(0.8983)

a = 63.94 degrees

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2 years ago

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3 years ago