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2 years ago

Find −115+(−35) Write your answer as a fraction in simplest form.

1 answer:

8 0

You would add the negatives and get a sum of:
-155.

to make it a fraction you just put it over one.

final answer:
-155/1

hope this helped!! xx

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Please help, thank you

Trava [24]

A=h•b/2
h=2x+1
b=2x
A=(2x+1) •2x/2
Divide numerator and denominator by 2
A=x(2x+1)

5 0

2 years ago

Find the area of the figure.

Ivanshal [37]

Step-by-step explanation:

Area of Rectangle at the top =

Length x Width

$= 3.5 \times 3 \\ = 10.5 {mm}^{2}$

Area of Quarter- Circle = 1/4 x Area of Circle

$= \frac{1}{4} \pi {r}^{2} \\ = \frac{1}{4} \pi( {3.5}^{2} ) \\ = 3 \frac{1}{16} \pi {mm}^{2}$

Area of Rectangle at the right = Length x Width

$= 5.5 \times 3.5 \\ = 19.25 {mm}^{2}$

Total Area = Area of 2 Rectangles + Area of Quarter Circle

$= 10.5 + 3 \frac{1}{16} \pi + 19.25 \\ = 29.75 + 3 \frac{1}{16}\pi \\ = 32 \frac{13}{16}\pi {mm}^{2}$

I will leave the answer in terms of pi as I'm not sure if you need to round off your answer.

8 0

2 years ago

Will give brainliest to right answer please help!! 15 points -A sample of a radioactive isotope had an initial mass of 370 mg

Alex777 [14]

Answer:

78 mg

Step-by-step explanation:

370 mg - 110 mg = 260 mg (amount decayed in the 8 total years)

260 mg ÷ 8 years = 32.5 mg decayed per year

32.5 × 9 years (years between 2008 and 2017) = 292.5 mg

370 mg - 292.5 mg = 77.5 mg, rounded to 78 mg

4 0

3 years ago

Read 2 more answers

What os the domain of the given function

kramer

I need to see the function In order to do this

6 0

3 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

3 years ago