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3 years ago

For all real numbers a and b, 2a • b = a2 + b2 True or False? explain please

Mathematics

1 answer:

vodka [1.7K]3 years ago

3 0

Answer:

True

Step-by-step explanation:

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-2(7-15) What is the value of 4 ? -4 -2 02. O 4

galben [10]

Answer:

Step-by-step explanation:

=-2(7-15)

USING 'BODMAS' RULE

= -2(-8)

= -2*-8

= 16

7 0

3 years ago

Solve for c c/3=30/5=

yaroslaw [1]

Answer:

= 18

Step-by-step explanation:

$\frac{c}{3}$ = $\frac{30}{5}$

Cross multiply

5 * c = 30 * 3

5c = 90

Divide both sides by 5

c = 90/5

c = 18

6 0

3 years ago

Read 2 more answers

What is the unit rate of $2.98

Semenov [28]

Answer:

Step-by-step explanation:

The existence of a unit rate implies that you have TWO variables. Here you list only one: $.

If you pay $2.98 for 6 units of something, such as pieces of fruit, then the unit rate would be:

$2.98

----------------------- = $0.50/piece

6 pieces of fruit

7 0

3 years ago

Find the formula for an exponential function that passes through the two points given.

goldenfox [79]

Answer:

$y = 7(3^x)$

Step-by-step explanation:

Given

$(x_1,y_1) = (0,7)$

$(x_2,y_2) = (4,567)$

Required

Determine the formula

An exponential function is of the form:

$y = ab^x$

For point $(x_1,y_1) = (0,7)$

$7 = ab^0$

$7 = a*1$

$7 = a$

$a = 7$

For point $(x_2,y_2) = (4,567)$

$567 = ab^4$

Substitute 7 for a

$567 = 7*b^4$

Divide both sides by 7

$81 = b^4$

Take 4th root of both sides

$\sqrt[4]{81} =\sqrt[4]{b^4}$

$\sqrt[4]{81} =b$

$b = \sqrt[4]{81}$

$b = 3$

Substitute 7 for a and 3 for b in $y = ab^x$

$y = 7 * 3^x$

$y = 7(3^x)$

3 0

3 years ago

Select True or False for each statement.

SSSSS [86.1K]

$\left( \dfrac 1 {64} \right)^{- 5/6} =64^{5/6} = (\sqrt[6]{64})^5 = 2^5 =32$

TRUE

$\sqrt[5]{36^4}=36^{4/5}$

which surely isn't 36. FALSE

$\sqrt{12} - \dfrac 2 5 \sqrt{75} = 2 \sqrt{3} -\dfrac 2 5 (5) \sqrt{3} = 0$

FALSE

For the fourth one we have a

$\sqrt{98b} + \sqrt{2b}$

which isnt

$10\sqrt{b}$

so this is FALSE.

$\dfrac{1}{(\sqrt 5 - \sqrt 6)^2}$

$= \dfrac{1}{(\sqrt 5 - \sqrt 6)^2} \cdot \dfrac{(\sqrt 5 + \sqrt 6)^2}{(\sqrt 5 + \sqrt 6)^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{ ( (\sqrt 5 - \sqrt 6)(\sqrt 5 + \sqrt 6))^2}$

$= \dfrac{(\sqrt 5 + \sqrt 6)^2}{( 5-6)^2}$

$=(\sqrt 5 + \sqrt 6)^2$

No fractions in that one so FALSE.

3 0

3 years ago