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3 years ago

I need question a and b done pls help

Mathematics

2 answers:

FrozenT [24]3 years ago

8 0

Answer:

A. 5

B. 12

Step-by-step explanation:

What you do is you substitute -2 into the a and the 3 into the b.

a. 3(-2)^2-5(3)+8 b. 2/3(3)-5(-2)

3(4)-15+8 2-(-10)

12-15+8 12 is your answer.

-3+8

5 is your answer.

CaHeK987 [17]3 years ago

4 0

For (a)

3a^2 - 5b + 8

Let a be -2 and b be 3.

3(-2)^2 - 5(3) + 8

3(4) - 15 + 8

12 - 15 + 8

-3 + 8

Answer: 5

For (b)

Let a be -2

Let b be 3

(2/3)b - 5a

(2/3)(3) - 5(-2)

2 + 10

Answer: 12

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Expanded form for 1.7

Verdich [7]

One and seven tenths

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yra solves the equation as shown. - 3(x+2) = 9 1. - 3x - 6 = 9 2. - 3x = 15 3. x = - 5 The property Tyra used in line 1 was the:

Molodets [167]

Answer:

Distributive property says that:

$a\cdot(b+c) = a\cdot b + a\cdot c$

Addition property of equality:

if x = y then x+z = y+z

Division property of equality:

if x = y then $\frac{x}{z}=\frac{y}{z}$

As per the statement:

Given the equation: $-3(x+2)=9$

Tyra solve this equation as shown below:

$-3x -6 = 9$ [Distributive property]

$-3x = 15$ [Addition property of equality]

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Therefore, the property Tyra used in line 1 was : Distributive Property.

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3 years ago

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Use the following image to assist in your answer. a = - 6 - 9 - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

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3 years ago

What is the length of the hypotenuse? If necessary, round to the nearest tenth.

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