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3 years ago

What are the x- and y-intercepts of f(x) = x2 - 49?

Mathematics

1 answer:

LenaWriter [7]3 years ago

7 0

Answer:

<h3> x-intercepts: (-7, 0) and (7, 0)</h3><h3> y-intercept: (0, -49)</h3>

Step-by-step explanation:

x-intercepts: f(x) = 0

x² - 49 = 0

x² = 49

x = -7 ∨ x = 7

Points: (-7, 0) and (7, 0)

y-intecept: y = f(0)

f(0) = 0² - 49 = -49

Point: (0, -49)

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Answer:

The required expression is $2(2x+\frac{5}{2})$

Step-by-step explanation:

Let the number be x

We are supposed to write the algebraic expression of given statement

Double the number = 2x

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Now we are given that Double the sum of a double the number and quotient of five and two

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So, Double the sum of a double the number and quotient of five and two = $2(2x+\frac{5}{2})$

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3 years ago

The value x = 3 is a solution to each of the following except which?

ololo11 [35]

Answer:

$x \: is \: not \: a \: solution \: in \to \: 5x + 4 < 19. \\$

Step-by-step explanation:

$\: if \: x = 3 \: then \\ 4x-1=2x +5 \: is \: same \: as \to \: \\ 4(3) - 1 = 2(3) + 5\\ 12 - 1 = 6 + 5 \\ 11 =11 \to \: \\ \boxed{ x \: is \:a \: solution \: here} \\ \\ if \: x = 3 \\ then \to \\ 5x + 4 < 19\: is \: same \:as \to \\ 5(3) + 4 < 19 \\ 15 + 4 < 19 \\ 19 \: is \: equal \: to \: 19 : \\ 19 \: can \: not \: be \: less \: than \: 19 \\ \boxed{ henc e\: x \: is \: not \: a \: solution \: here}$

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Step-by-step explanation:

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2 years ago

If the product of two positive fractions a and b is 15/56, find three pairs of possible values for a and b

Andrews [41]

Answer:

First possible pairs = (3/7)*(5/8) = 15/56

second possible pairs = (3/4)*(5/14) = 15/56

third possible pairs = (1/2)*(15/28) = 15/56

Step-by-step explanation:

According to the question,

a x b = 15/56

three possible pairs will be as follows:

Before proceeding to make possible pairs, we have to know the prime factorization of the numbers -

15 = 1, 3, 5, 15

56 = 1, 2, 4, 7, 8, 14, 28, 56

As we need three possible pairs, we have to check which pairs take us to 15/56. Again, since the fraction is positive, therefore, the fractions will be proper. Therefore, 3/2, or 5/4 will not be counted. Therefore,

First possible pairs = (3/7)*(5/8) = 15/56

second possible pairs = (3/4)*(5/14) = 15/56

third possible pairs = (1/2)*(15/28) = 15/56

fourth possible pairs = (3/8)*(5/7) = 15/56

So, we can get those four possible pairs. Among those, first 3 pairs are different. Therefore, those are possible pairs.

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3 years ago