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3 years ago

Question 9) Help me please

Mathematics

2 answers:

Rufina [12.5K]3 years ago

8 0

C. (-9, -2)

Hope this helps!

marta [7]3 years ago

4 0

Answer:

the pidodoesnt open for me

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Joanna split 3 pitchers of water equally among her 8 plants. What fraction of a pitcher did each plant get?

MariettaO [177]

The answer is 3/8 of a pitcher

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3 years ago

Give an example of an infinite set.

Montano1993 [528]

Infinite sets may be countable or uncountable. Some examples are: the set of all integers, {..., -1, 0, 1, 2, ...}, is a countably infinite set; and. the set of all real numbers is an uncountably infinite set.

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3 years ago

Read 2 more answers

When 23 is subtracted from 5 times a certain number the result is 87

allsm [11]

5x-23=87
add 23 to both sides
5x=110
divide both sides by 5
x=22

To check your answer...
5(22)-23=87
110-23=87
87=87

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4 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

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3 years ago

Megan has a bag of beads. She uses 28 beads to make each necklace. She makes 12 necklaces. Megan has 135 beads left over.

Bezzdna [24]

If you are asking for how many beads she used in total for the 12 necklaces it is
336. If you are asking how many beads she had before making the necklaces it is 471. If you are asking how many necklaces can she make with the leftover beads it is about 4. Hope this helps plz mark a crown.

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3 years ago