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3 years ago

Please help me with this math problem!! :)

Mathematics

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

x= 4

(not sure about a, sorry friend)

Step-by-step explanation:

Since the shape is congruent with the other sides it's the same.

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Which fractions are equal to −9/36?

STatiana [176]

Answer:

1/4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The populations P (in thousands) of a certain city from 2000 through 2007 can be modeled by P = 1674.7e^kt, where t represents t

Makovka662 [10]

Answer:

k = 0.02278

Step-by-step explanation:

Given that:

The population P = 1674.7e^kt

where;

t represents the year,

if year 2000 represents t = 0

Then, the year 2007 represents = 7; &

year 2006 represents = 6

In 2006, population recorded in the city was about 1,920 thousand

Therefore;

$1920 = 1674.7e^{(k\times 6)$

Taking the logarithm of both sides, we get:

$\mathtt{ In}(1920) = \mathtt{ In} ( 1674.7)+ 6k$

7.560080465= 7.423389324 + 6k

7.560080465 - 7.423389324 = 6k

0.136691141 = 6k

k = 0.136691141/6

k ≅ 0.02278

5 0

3 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

4 years ago

Given: ΔPSQ, PS = SQ

horrorfan [7]

<h2>Answer:</h2>

To solve this problem we will use Heron's formula:

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where $a, \ b \ and \ c$ are the side lengths of the triangle and $s$ is the semiperimeter (half the perimeter of the triangle). We know that:

$Perimeter \ P=\triangle PSQ=PS+PQ+SQ: \\ \\ \triangle PSQ=P=50 \\ \\ Semiperimeter \ s: \\ \\ s=\frac{P}{2}=25$

Also:

$(I) \ PS=SQ \\ \\ (II) \ SQ-PQ = 1 \\ \\ (III) \ PS+PQ+SQ=50 \\ \\ \\ (I) \ into \ (III): \\ \\ SQ+PQ+SQ=50 \\ \\ \therefore (IV) \ 2SQ+PQ=50 \\ \\ From \ (II): \\ \\ PQ=SQ-1 \\ \\ (II) \ into \ (IV): \\ \\ 2SQ+(SQ-1)=50 \\ 3SQ-1=50 \\ 3SQ=51 \\ \\ \boxed{SQ=17} \\ \\ \boxed{PS=17} \\ \\ PQ=SQ-1=17-1 \therefore \boxed{PQ=16}$

Finally:

$A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ A=\sqrt{s(s-PS)(s-SQ)(s-PQ)} \\ \\ A=\sqrt{s(s-17)(s-17)(s-16)} \\ \\ A=\sqrt{25(25-17)(25-17)(25-16)} \\ \\ \boxed{A=120}$