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3 years ago

Write the expression in standard form. six divided by quantity four minus thirteen i.

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Answer:

$\frac{6}{4} - 13i$

Step-by-step explanation:

six divided by quantity four minus thirteen i

Standard form of the expression is a+bi

six divided by quantity four means we divide 6 and 4

six divided by quantity four is $\frac{6}{4}$

Now minus thirteen i

So we subtract 13i from the fraction

$\frac{6}{4} - 13i$

It is in the form of a+bi

dangina [55]3 years ago

4 0

6 $\frac{6}{4-13i}$

I'm not sure if the i is supposed to be there or not. Is it a variable that is connected to 13?

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3 years ago

A photo album costs $10.49 before discount and tax.

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Answer:

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Step-by-step explanation:

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3 years ago

a jeweler has 36 inches of silver chain She needs 5 times that much to make some necklaces and. 3 times that amount to make some

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Answer:

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Step-by-step explanation:

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2 years ago

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1. This data shows the pocket money that year 9 pupils receive each

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3 years ago

Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM

denis23 [38]

Answer:

The length of side KM is $\sqrt{109}$ units.

Step-by-step explanation:

Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=3:5, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, $ME=3\sqrt{5}$ .

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is $3\sqrt{5}$ .

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since LM:KN=3:5, therefore LM is 3x and KN is 5x.

$\frac{3x+5x}{2}=8$

$\frac{8x}{2}=8$

$x=2$

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

$KN=KA+AE+EN$

$10=6+2(EN)$ (KA=EN, isosceles trapezoid)

$EN=2$

$KE=KN-EN=10-2=8$

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Therefore the length of side KM is $\sqrt{109}$ units.

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3 years ago

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