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3 years ago

9

Find 3 consecutive integets such that the sum of the first two integers is 24 more than the third.

1 answer:

Elina [12.6K]3 years ago

4 0

Answer:

breath

Step-by-step explanation:

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Find the missing measurement for each figure.

earnstyle [38]

To find h, divide the area by the base:

h = 2442 / 12 = 203.5 km.

6 0

3 years ago

I'll mark the best or the first as brainliest.

FinnZ [79.3K]

Answer:

<em><u>Here</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>error</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em>

<em><u> $\pi {r}^{2}$ </u></em>

<em><u>It</u></em><em><u> </u></em><em><u>should</u></em><em><u> </u></em><em><u>have</u></em><em><u> </u></em><em><u>been</u></em><em><u> </u></em><em><u>,</u></em>

<em><u> $2\pi {r}^{2}$ </u></em>

Step-by-step explanation:

<em><u>Corrected</u></em><em><u> </u></em><em><u>formula</u></em><em><u>:</u></em>

<em><u> $2\pi {r}^{2} + 2\pi rh$ </u></em>

<em><u> $= 2\pi ({5}^{2}) + 2\pi(5)(10.6)$ </u></em>

<em><u> $= 50\pi + 106\pi$ </u></em>

<em><u> $= 156\pi$ </u></em>

<em><u> $= 489.84 {yd}^{2}$ </u></em>

<em><u> $approx . \: 489.8 {yd}^{2} (ans)$ </u></em>

6 0

3 years ago

A satellite television company charges a one-time installation fee and a monthly service charge. The total cost is modeled by th

Marat540 [252]

<span>(2) y is the total cost, x is the number of months of service, $40 is the installation fee, and $90 is the service charge per month.</span>
y = 40 + 90x

y = total cost
x = number of months of service
40 = one time installation fee
90 = monthly service charge.

Let us assume that the number of months of service is 10 months. then, the total cost would be:

y = 40 + 90(10)
y = 40 + 900
y = 940

5 0

3 years ago

Help me please !!<br> I need this ASAP please helpppp

Kobotan [32]

Answer:

C

Step-by-step explanation:

$y=(2x-3)/5$

$5y=2x-3$

$(5y+3)/2=x$

$(5x+3)/2=y$

$f^{-1}(x)=(5x+3)/2$

8 0

3 years ago

2x2 – 9x + 2 = –1

Reika [66]

Answer:

C) The discriminant is greater than 0, so there are two real roots

Step-by-step explanation:

<u>Discriminant</u>

$b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0$

$\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}$

$\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}$

$\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}$

$2x^2-9x+2=-1$

$\implies 2x^2-9x+3=0$

$\implies a=2, b=-9, c=3$

$\begin{aligned}b^2-4ac &=(-9)^2-4(2)(3)\\& =81-24\\&=57 > 0\implies \textsf{two real roots}\end{aligned}$

5 0

2 years ago