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stiks02 [169]
3 years ago
12

A cross between two heterozygous pea plants with yellow seeds produced 1,719 yellow seeds and 573 green seeds. What is the ratio

of yellow to green seeds?
A cross between two heterozygous pea plants with smooth green pea pods produced 87 bumpy yellow pea pods, 261 smooth yellow pea pods, 261 bumpy green pea pods, and 783 smooth green pea pods. What is the ratio of bumpy yellow to smooth yellow to bumpy green to smooth green pea pods?
Mathematics
2 answers:
Nutka1998 [239]3 years ago
8 0

Answer:

hi

Step-by-step explanation:

11Alexandr11 [23.1K]3 years ago
4 0
1). 3:1

2). 1:3 and 1:2





Hope this helps :)
You might be interested in
1) Write an equation that includes a variable to show how much money matt started with
12345 [234]
1) m-13=7

2)m-13=7
m-13+13=7+13
m=20
so he had $20 he spent $13 so $7 was left!


BRAINLIEST???
7 0
3 years ago
Heather works in an appliance store. She earns $1,500 per month plus a commission of 4% on her monthly sales. Her weekly sales a
Nataly_w [17]

Answer: $1980

Step-by-step explanation:

First, let’s find out her monthly sales by adding up the 4 weekly sales.

2400+ 3200 + 5200 + 1200 = 12000

Next, find the amount of commission on 12000

4%/100=0.04

12000*0.04=480

Next add her commission to her monthly wage of 1500

1500+480=1980

3 0
3 years ago
One pencil weigh fifteen grams. What is the weight of eight pencils?
artcher [175]

Answer:

120 grams

if they are all the same size

8 0
2 years ago
Express the product of z1 and z2 in standard form given that <img src="https://tex.z-dn.net/?f=z_%7B1%7D%20%3D%20-3%5Bcos%28%5Cf
Marta_Voda [28]

Answer:

Solution : 6 + 6i

Step-by-step explanation:

-3\left[\cos \left(\frac{-\pi }{4})\right+i\sin \left(\frac{-\pi }{4}\right)\right]\cdot \:2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi }{2}\right)\right]

This is the expression we have to solve for. Now normally we could directly apply trivial identities and convert this into standard complex form, but as the expression is too large, it would be easier to convert into trigonometric form first ----- ( 1 )

( Multiply both expressions )

-6\sqrt{2}\left[\cos \left(\frac{-\pi }{4}+\frac{-\pi \:\:\:}{2}\right)+i\sin \left(\frac{-\pi \:}{4}+\frac{-\pi \:\:}{2}\right)\right]

( Simplify \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right) for both \cos \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right) and i\sin \left(\frac{-\pi }{4}+\frac{-\pi }{2}\right) )

\left(\frac{-\pi }{4}+\frac{-\pi }{2}\right) = \left(-\frac{3\pi }{4}\right)

( Substitute )

-6\sqrt{2}\left(\cos \left(-\frac{3\pi }{4}\right)+i\sin \left(-\frac{3\pi }{4}\right)\right)

Now that we have this in trigonometric form, let's convert into standard form by applying the following identities ----- ( 2 )

sin(π / 4) = √2 / 2 = cos(π / 4)

( Substitute )

-6\sqrt{2}\left(-\sqrt{2} / 2 -i\sqrt{2} / 2 )

= -6\sqrt{2}\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right) = -\frac{\left(-\sqrt{2}-\sqrt{2}i\right)\cdot \:6\sqrt{2}}{2}

= -3\sqrt{2}\left(-\sqrt{2}-\sqrt{2}i\right) = -3\sqrt{2}\left(-\sqrt{2}\right)-\left(-3\sqrt{2}\right)\sqrt{2}i

= 3\sqrt{2}\sqrt{2}+3\sqrt{2}\sqrt{2}i:\quad 6+6i - Therefore our solution is option a.

4 0
4 years ago
Find the missing number. ( Show your work)
aleksandr82 [10.1K]

Step-by-step explanation:

For problem 1,

Let the required number be x,

2 x/4 = 11/4

or, (8+x)/4 = 11/4

or, 8+x = 44/4

or, 8+x = 11

or, x = 11-8

so, x = 3

For problem 2,

Let the required number be y,

y/10 = 4/5

or, 5y = 40

or, y = 40/5

so, y = 8

7 0
3 years ago
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