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3 years ago

14

What is the area of the parallelogram? cm2 7 cm 1.2 cm

2 answers:

9966 [12]3 years ago

7 0

What are numbers used though? I can’t answer if you don’t say the full problem because there aren’t any units or numbers.

KiRa [710]3 years ago

6 0

Do you have an image?

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Does anyone know how to do this please help me

Anit [1.1K]

Answer:

162°

Step-by-step explanation:

a circle has 360°, or the angle of a whole circle is that

so 45% of 360 =

0.45 times 360 = 162

7 0

3 years ago

What is the exact value of tan(300°)?

S_A_V [24]

Answer:

The exact value of tan 300° is ⏩ - √3

Hope it will help :)❤

7 0

3 years ago

Will give brainliest if answer correct// if p(x)= 3x^6 - 5x^5 - x + 7 , then p(2) equals: ??

djverab [1.8K]

$\\ \tt\hookrightarrow p(2)=3(2)^6-5(2)^5-2+7$

$\\ \tt\hookrightarrow p(2)=3(64)-5(32)+5$

$\\ \tt\hookrightarrow p(2)=192-160+5$

$\\ \tt\hookrightarrow p(2)=32+5$

$\\ \tt\hookrightarrow p(2)=37$

Option B is correct

4 0

3 years ago

Read 2 more answers

the area of maddies square garden is 450 square feet. she has purchased 92 feet of fence. is this the right amount of fence for

Keith_Richards [23]

Answer:

Not the right amount.

Step-by-step explanation:

A square has 4 sides so 92/4 to divide it evenly is 23.

23 times 23 for the are is 529 which is not equal to 450.

3 0

3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago