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valentina_108 [34]
3 years ago
11

PLZ help I don't understand!!!!

Mathematics
1 answer:
WITCHER [35]3 years ago
7 0

Answer:

5

Step-by-step explanation:

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How many times larger is 4×10^12 than 8×10^7?<br><br>2×10^4<br>2×10^5<br>5×10^4<br>5×10^5​
Leno4ka [110]
The answer is 5×10^4
6 0
3 years ago
What information do i need to solve word problems that involve measurement
forsale [732]
I think a ruler might help but you did not say what kind of measurement 
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4 years ago
Which of the following shows the factors of 5x2 + 17x + 6? A. (5x + 1)(x + 6) B. (5x + 2)(x + 3) C. (5x + 3)(x + 2) D. (5x + 6)(
zysi [14]

Answer:

B. (x+3)(5x+2)

Step-by-step explanation:

The given quadratic trinomial is;

5x^2+17x+6

Comparing this to ax^2+bx+c, we have a=5,b=17,c=6.

ac=5\times6=30

Two factors of 30 that add up to 17 is 15  and 2.

We split the middle term to obtain;

5x^2+15x+2x+6

We factor by grouping;

5x(x+3)+2(x+3)

Factor further

(x+3)(5x+2)

The correct answer is B.

5 0
4 years ago
Read 2 more answers
F(x)=x-4 how do u do this
irina1246 [14]
You were probably given a number to plug into the equation , which is no different then it being y=x-4 , just solve for the f(x)
4 0
3 years ago
Find the area of each parallelogram. What is the relationship between the areas?
castortr0y [4]

Answer:

Area of parallelogram is given by:

A = bh

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12 units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12 units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = QR \cdot PT

⇒Area of parallelogram TQRS = 12 \cdot 12 = 144 unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3 units

P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3 units

Then;

Area of parallelogram T'Q'R'S' = Q'R' \cdot P'T'

Area of parallelogram T'Q'R'S' = 9 \cdot 9= 81 unit square.

Now, we have to find the relationship between the areas.

\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}



7 0
3 years ago
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