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2 years ago

How do you use an exponent to be present a number such as 16

1 answer:

6 0

Maybe this can help you

This sounds like you're asking about a binary floating
point format, though I'm not positive because of the 
vocabulary you use. I interpret "characteristic" to mean 
"exponent" and "significand" as "fraction" (or "mantissa"). 

If that's true, then you've described: 

sign(1) + binary fraction(20) + binary exponent(10) 

Note that there are other possible interpretations (for 
example, a base-16 exponent). 

Since I'm assuming a binary exponent, we can use the same 
trick that IEEE 754 floating point standard does, and assume 
an "implicit" leading "1" to the fraction portion. This 
trick means that the fractional part will be between 
the following numbers (in base 2): 

1.00000000000000000000 (1 followed by twenty zeros)

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Simplify (−4.5)(−6)(5.4).

Fynjy0 [20]

The answer to (-4.5)(-6)(5.4)= 137.7

4 0

2 years ago

Johan can drive 270 miles on 9 gallons of gas. What is his average mileage?

VladimirAG [237]

Hey there!

The correct answer is 30 miles per gallon.

If he can drive 270 miles on 9 gallons, we want to find the unit rate, or miles per gallon. We can do this by dividing 270 by 9. To make it easier, we will take off the zero, and do 27/9, which is 3. Now we add the 0 back on, which gives us 30. Therefore, his average mileage is 30 miles per gallon.

Have a terrificly amazing day! :D

4 0

2 years ago

Read 2 more answers

Solve the following system by the comparison method.

agasfer [191]

Answer:

2p + q = 1

9p + 3q + 3 = 0

q = 1 - 2p

replace q = 1 - 2p into 9p + 3q + 3 = 0

9p + 3(1 - 2p) + 3 = 0

9p + 3 - 6p + 3 = 0

3p + 6 =0

3p = -6

p = -2

q = 1 - 2p

q = 1 -2(-2)

q = 1 + 4

q = 5

(-2 , 5)

there for the answer would be

{(-2, 5)}

Hopefully this was helpful <3 :3

5 0

2 years ago

13 POINTS- please help me

allsm [11]

Answer:

See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures $(6x+20)^{\circ}$ and $(x+100)^{\circ}$ are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

$6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16$

Then

$(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}$

17. Two parallel lines are cut by transversal. Angles with measures $(2x+12)^{\circ}$ and $(3x-22)^{\circ}$ are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

$2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34$

Then

$(2x+12)^{\circ}=(2\cdot 34+12)^{\circ}=80^{\circ}\\ \\(3x-22)^{\circ}=(3\cdot 34-22)^{\circ}=80^{\circ}$

18. Two parallel lines are cut by transversal. Angles with measures $(6x-7)^{\circ}$ and $(5x+10)^{\circ}$ are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

$6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17$

Then

$(6x-7)^{\circ}=(6\cdot 17-7)^{\circ}=95^{\circ}\\ \\(5x+10)^{\circ}=(5\cdot 17+10)^{\circ}=95^{\circ}$

19. The diagram shows two complementary angles with measures $2x^{\circ}$ and $56^{\circ}$ . The measures of complementary angles add up to $90^{\circ},$ then

$2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17$

Hence,

$2x^{\circ}=2\cdot 17^{\circ}=34^{\circ}$

Check:

$34^{\circ}+56^{\circ}=90^{\circ}$

20. Angles $\angle 1$ and $\angle 2$ are vertical angles. By vertical angles theorem, vertical angles are congruent, so

$m\angle 1=m\angle 2\\ \\5x+7=3x+15\\ \\5x-3x=15-7\\ \\2x=8\\ \\x=4$

Hence,

$m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}$

21. $\angle 5$ and $\angle 8$ are supplementary. The measures of supplementary angles add up to $180^{\circ},$ so

$m\angle 5+m\angle 8=180^{\circ}\\ \\3x-40+7x-120=180\\ \\10x-160=180\\ \\10x=180+160\\ \\10x=340\\ \\x=34$

Therefore,

$m\angle 5=(3x-40)^{\circ}=(3\cdot 34-40)^{\circ}=62^{\circ}\\ \\m\angle 8=(7x-120)^{\circ}=(7\cdot 34-120)^{\circ}=118^{\circ}\\ \\62^{\circ}+118^{\circ}=180^{\circ}$

7 0

2 years ago

Solve the absolute value inequality: |x + 12| + 5 < 27 Isolate the absolute value by subtracting 5 from both sides.

Feliz [49]

Answer:

- 34 < x < 10

Step-by-step explanation:

Inequalities of the type | x | < a, always have a solution of the form

- a < x < a

given | x + 12 | + 5 < 27 ( subtract 5 from both sides )

| x + 12 | < 22, then

- 22 < x + 12 < 22 ( subtract 12 from all 3 intervals )

- 34 < x < 10

8 0

3 years ago

Read 2 more answers