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3 years ago

9-17 = 23 q= what does q equal?

Mathematics

1 answer:

UkoKoshka [18]3 years ago

3 0

Answer:

q = - 14 / 17

Step-by-step explanation:

9 - 17q = 23

-17q = 23 - 9

-17 q = 14

q = - 14 / 17

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Please help!! Very urgent..thank you!!

erastova [34]

Answer : 4 times

Here it's given that ,

The height and base of the butterfly sitting on the stem (red butterfly) is two times greater than the height and base of the butterfly sitting on the flower .

And we need to find out how many times the area of red winged butterfly is greater than that of sitting on the flower (blue butterfly) .

Let us take ,

base of blue butterfly be b
height of blue butterfly be h
Area be A .

Then ,

base of red butterfly will be 2b .
height of red butterfly will be 2h .
Area be A' .

We know that ,

→ area of the triangle = 1/2 × base × height

So that ,

→ A/A' = (1/2 * b * h) ÷ (1/2 *2b *2h)

→ A/A' = bh/4bh

→ A/A' = 1/4

→ A' = 4A

Henceforth the area of blue butterfly is 4 times greater than that of red winged butterfly .

I hope this helps.

6 0

2 years ago

Help quick, it's based on polynomials . Find the area of the shape

IrinaK [193]

Answer:

Step-by-step explanation:

I got you first you want to get some little ceasers

7 0

2 years ago

#1 - HINT - you are finding the total amount make sure you add the interest you calculate to the original deposit amount

Andreyy89

Answer:

$292.50

Step-by-step explanation:

Equation - A = P(1 + rt)

Calculation:

First, converting R percent to r a decimal

r = R/100 = 2.5%/100 = 0.025 per year.

Solving our equation:

A = 260(1 + (0.025 × 5)) = 292.5

A = $292.50

The total amount accrued, principal plus interest, from simple interest on a principal of $260.00 at a rate of 2.5% per year for 5 years is $292.50.

8 0

3 years ago

Read 2 more answers

Find the circumference of this circle.

finlep [7]

Answer:

81.64

Step-by-step explanation:

2(3.14)=6.28

6.28*13=81.64

:))))

3 0

3 years ago

Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0

marin [14]

If you do in fact mean $f(1)=f(6)=0$ (as opposed to one of these being the derivative of $f$ at some point), then integrating twice gives

$f''(x) = -\dfrac1{x^2}$

$f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1$

$f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2$

From the initial conditions, we find

$f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0$

$f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)$

Eliminating $C_2$ , we get

$(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))$

$-5C_1 = \ln(6)$

$C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)$

Then

$\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}$

3 0

2 years ago