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2 years ago

Calculate the length of AC to one decimal place in the trapezium below.

Mathematics

1 answer:

a_sh-v [17]2 years ago

7 0

Answer:

First we will split trapezium on two geometric figure.

One is rectangle and the second is right triangle.

When we subtract 11-4=7cm we get one cathetus of the right triangle a=7cm also we know hypotenuse which is c=16cm.

We will use Pythagorean theorem to find the second cathetus b

b=√16∧2-7∧2= √256-49= √207 ≈ 14.39cm =DC =>

AC =√(DC)∧2+(AD)∧2= √14.39∧2+11∧2= √207+121= √328= 18.1cm

AC=18.1cm

Good luck!!

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Complete the equation describing how x and y are related. Help please!!!

Ivahew [28]

Answer:

-4

Step-by-step explanation:

Y = -4x+-2

Not quite sure how to explain it on here but substitute it and it works. :)

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2 years ago

The function y = square root sign x is translated using the rule (x, y) → (x – 7, y + 2)to create f(x). What is the domain of f(

Mama L [17]

Answer:

Second option {x | x > 7}

Step-by-step explanation:

We have the function

$y = \sqrt{x}$

We know that the square root of a negative number has no solution in the real ones. Therefore the domain of this function is $x > 0$

When applying the transformation:

$(x, y) \to (x - 7, y + 2)$ we have a translation of the original function in 7 units to the right and 2 units to the top:

$f(x) = \sqrt{x-7} + 2$

In the same way we must guarantee that $(x-7)> 0$

Then $x > 7$ .

Therefore the domain of f(x) is {x | x > 7}

7 0

2 years ago

Read 2 more answers

Which equation describes this line?

Anettt [7]

The last equation would best describe the line.

y-4=3(x+2)

y-4=3x + 6

y=3x + 10

4 0

2 years ago

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Hiro has a stack of cards with one number from the set 1, 1, 2, 2, 3, 3, 3, 4 written on each card. What is the probability that

professor190 [17]

Answer:

Therefore, the probability is P=3/32.

Step-by-step explanation:

We know that Hiro has a stack of cards with one number from the set 1, 1, 2, 2, 3, 3, 3, 4 written on each card.

We calculate the probability that he pulls out a 3 first and then pulls out a 2 without replacing them.

The probability that he pulls out a 3 first is 3/8.

The probability of a second card being 2 is 2/8.

We get:

$P=\frac{3}{8}\cdot \frac{2}{8}\\\\P=\frac{6}{64}\\\\P=\frac{3}{32}$

Therefore, the probability is P=3/32.

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2 years ago

Colin invests £2350 into a savings account. The bank gives 4.2% compound interest for the first 4 years and 4.9% thereafter. How

mixer [17]

To solve this, we are going to use the compound interest formula: $A=P(1+ \frac{r}{n} )^{nt}$
where
$A$ is the final amount after $t$ years
$P$ is the initial investment
$r$ is the interest rate in decimal form
$n$ is the number of times the interest is compounded per year

For the first 4 years we know that: $P=2350$ , $r= \frac{4.2}{100} =0.042$ , $t=4$ , and since the problem is not specifying how often the interest is communed, we are going to assume it is compounded annually; therefore, $n=1$ . Lest replace those values in our formula:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2350(1+ \frac{0.042}{1} )^{(1)(4)}$
$A=2350(1+0.042)^{4}$
$A=2770.38$

Now, for the next 6 years the intial investment will be the final amount from our previous step, so $P=2770.38$ . We also know that: $r= \frac{4.9}{100} =0.049$ , $t=6$ , and $n=1$ . Lets replace those values in our formula one more time:
$A=P(1+ \frac{r}{n} )^{nt}$
$A=2770.38(1+ \frac{0.049}{1})^{(1)(6)$
$A=2770.38(1+0.049)^6$
$A=3691.41$

We can conclude that Collin will have <span>£3691.41 in his account after 10 years.</span>

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2 years ago