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2 years ago

Find area of parallelogram multiple choice

Mathematics

1 answer:

alina1380 [7]2 years ago

4 0

In ∆ABD

$\\ \sf\longmapsto Area=\dfrac{1}{2}bh=\dfrac{1}{2}(7\times 4)$

In ∆BCD

$\\ \sf\longmapsto Area=\dfrac{1}{2}(7\times 4)$

Now

IN PARALLELOGRAM ABCD

$\\ \sf\longmapsto Area=\dfrac{1}{2}(7\times 4)+\dfrac{1}{2}(7\times 4)$

Option A

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Each equation represents a linear relationship. Examine each and determine the slope and y- intercept

marysya [2.9K]

Answer:

Slope is -3/7. Y intercept is 250/7.

Step-by-step explanation:

Since they asking for the slope and y intercept, let put it in slope intercept form.

$y = mx + b$

where m is the slope and b is the y intercept.

The equation given is in standard form so we will convert it to slope intercept.

$3x + 7y = 250$

$7y = 250 - 3x$

$y = \frac{250}{7} - \frac{3}{7} x$

$y = - \frac{3}{7} x + \frac{250}{7}$

The slope is -3/7

The y intercept is 250/7

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3 years ago

K=1/2mv squared <br> Rearrange the formula to solve for m

musickatia [10]

Answer:

Step-by-step explanation:

1/2mv = k

rearranged can be:

m = k / 1/2 over v

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3 years ago

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allochka39001 [22]

Answer:

b. 82m³

Step-by-step explanation:

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3 years ago

Read 2 more answers

A worker was paid a salary of $10,500 in 1985. Each year, a salary increase of 6% of the previous year's salary was awarded. How

Mazyrski [523]

Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.

Here is how much the worker earned each year:

In the year 1985 the worker earned <span>$10,500.

</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:

$10,500(1+0.06).

In the year 1987 the worker earned

$10,500(1+0.06) + 0.06[$10,500(1+0.06)].

Now we can factorize $10,500(1+0.06) and write the earnings as:

$10,500(1+0.06) [1+0.06]= $$10,500(1.06)^2$ .

Similarly we can check that in the year 1987 the worker earned $$10,500(1.06)^3$ , which makes the pattern clear.

We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:

$10,500+10,500(1.06)^1+10,500(1.06)^2+10,500(1.06)^3...10,500(1.06)^{26}$ .

Factorizing, we have

$=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]$

We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula

$\sum_{i=1}^{n} a_i= a(\frac{1-r^n}{1-r})$ (where a is the first term and r is the common ratio) we have:

$\sum_{i=1}^{26} a_i= 1(\frac{1-(1.06)^{26}}{1-1.06})= \frac{1-4.55}{-0.06}= 59.17$ .

Finally, multiplying 10,500 by 59.17 we have 621.285 ($).

The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating $(1.06)^{26}$ .

Answer: D

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3 years ago

A and B are points on a line that has the slope 2. The abscissa of B exceeds the abscissa of A by 3.5. State the relationship of

Troyanec [42]

Answer:

The ordinate of B exceeds the ordinate of A by 7

Step-by-step explanation:

Let

A(x1,y1),B(x2,y2)

where

x1 is the abscissa of A

x2 is the abscissa of B