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3 years ago

Which two integers is 42 between? -8 and -7 -5 and -4 -7 and -6 -9 and -8

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

3 0

Answer:

6 and 7

Step-by-step explanation:

we know that 42 will lie between 36 and 49

And 36=6; 49=7

so we can be sure that 42 lies between 6 and 7

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The coordinates of CD are C (-7,-2) and D (-3,-5). Point E is the midpoint of CD. What are the coordinates of the midpoint E?

goldenfox [79]

The coordinates of the midpoint of CD with endpoint at C(-7,-2) and D(-3,-5) is E(-5, -3.5)

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Let (x, y) represent the coordinate of the midpoint E. Point E is the midpoint of CD, hence:

x = [(-7) + (-3)] / 2 = -5

y = [(-2) + (-5)] / 2 = -3.5

The coordinates of the midpoint of CD with endpoint at C(-7,-2) and D(-3,-5) is E(-5, -3.5)

Find out more on equation at: brainly.com/question/2972832

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2 years ago

Which equation is the inverse of 2(x - 2)2 = 8(7 + y)?

Scrat [10]

The inverse of an equation is the opposite of the equation

The inverse of the function is $y =2 \pm \sqrt{28 + 4x$

<h3>How to determine the inverse of the function </h3>

The equation is given as:

$2(x - 2)^2 = 8(7 + y)$

Divide both sides of the equation by 2

$(x - 2)^2 = 4(7 + y)$

Open the bracket

$(x - 2)^2 = 28 + 4y$

Take the square roots of both sides

$x - 2 = \pm\sqrt{28 + 4y$

Add 2 to both sides

$x =2 \pm \sqrt{28 + 4y$

Swap the occurrences of x and y

$y =2 \pm \sqrt{28 + 4x$

Hence, the inverse of the function is $y =2 \pm \sqrt{28 + 4x$

Read more about inverse functions at:

brainly.com/question/8120556

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2 years ago

the larger of two numbers is 3 more than the smaller number if the sum of the numbers is 73 what are the two numbers

Zinaida [17]

The two numbers used in the equation are 70+3. :D

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3 years ago

Calculate a vector c in the same direction of the vector a and that has the same length as the vector b

Viktor [21]

For any scalar $t>0$ , $t\vec a$ is a vector pointing in the same direction as $a$ .

To force this vector to have the same length as another vector $\vec b$ , first normalize $t\vec a$ by dividing it by its norm,

$\dfrac{t\vec a}{\|t\vec a\|}=\dfrac{t\vec a}{t\|\vec a\|}=\dfrac{\vec a}{\|\vec a\|}$

then multiply this vector by the norm of $\vec b$ , so that

$\vec c=\dfrac{\|\vec b\|}{\|\vec a\|}\vec a$

To confirm that $\vec c$ has the same length as $\vec b$ :

$\|\vec c\|=\left\|\dfrac{\|\vec b\|}{\|\vec a\|}\vec a\right\|=\dfrac{\|\vec b\|}{\|\vec a\|}\|\vec a\|=\|\vec b\|$

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3 years ago

What is another way to write 15 minutes before 9?

dimulka [17.4K]

Answer:

8:45

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