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ss7ja [257]
3 years ago
11

Which of the following could not represent the y-intercept of a Direct Variation graph?

Mathematics
2 answers:
ipn [44]3 years ago
6 0
It’s C Bc that would b the outcome of the algorithm w x in the mix
ZanzabumX [31]3 years ago
4 0
C
Sorry if I am wrong
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Help please ..............
Cloud [144]

Answer:

x = 437.3 ft

Step-by-step explanation:

The angle at the top of the triangle = 90° - 29° = 61°

Using the sine ratio in the right triangle

sin61° = \frac{opposite}{hypotenuse} = \frac{x}{500}

Multiply both sides by 500

500 × sin61° = x, hence

x ≈ 437.3

8 0
3 years ago
Read 2 more answers
Solve each system of linear equations by elimination. <br><br>2v=-5 +10 <br>-15=-3x + 5y​
Burka [1]

Answer:Just solve it. This is Algebra. Not that hard. Use Substation Method

Step-by-step explanation:

3 0
3 years ago
Fruit is being handed out at a picnic. The basket contains 5 bananas, 3 oranges and 2 apples. If two pieces of fruit are handed
stepan [7]
Because there are 9 fruits, it should be n/9, depending on the fruit.
Apples are 2/9, so:
The combination for apples would have a probability of 2/9<span />
7 0
3 years ago
Which polynomial function has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1? f(x) = (x – 2i)(x – 3i) f(x) =
kykrilka [37]

Answer:

P(x)=(x-2i)(x-3i)

Step-by-step explanation:

Build a Polynomial Knowing its Roots

If we know a polynomial has roots x1, x2, ..., xn, then it can be expressed as:

P(x)=a(x-x1)(x-x2)...(x-xn)

Where a is the leading coefficient.

Note the roots appear with their signs changed in the polynomial.

If the polynomial has a leading coefficient of 1 and roots 2i and 3i with multiplicity 1, then:

P(x)=1(x-2i)(x-3i)

\mathbf{P(x)=(x-2i)(x-3i)}

8 0
3 years ago
Read 2 more answers
Tyler is offered to choose between three boxes that look identical. The first box contains 1 red pills and 19 blue pills, the se
Dafna1 [17]

Answer:

A) 0.15

B) 0.36

Step-by-step explanation:

First Box consist 1 red pills and 19 blue pills

second box consist 6 red pills and 14 blue pills

Third box consist 2 red pills and 18 blue pills

a) Prob ( randomly chosen pill is red) = \frac{1}{3} \times \frac{1}{20} + \frac{1}{3} \times \frac{6}{20} + \frac{1}{3} \times \frac{2}{20}

\frac{1}{60} (1+6+2) = 0.15

b)Prob ( 1 reds and 10 blue pilss) =\frac{1}{3} \frac{^1C_1\times ^{19}C_{10}}{^{20}C_{11}} + \frac{1}{3} \frac{^6C_1\times ^{14}C_{10}}{^{20}C_{11}} + \frac{1}{3} \frac{^2C_1\times ^{18}C_{10}}{^{20}C_{11}}

= \frac{1}{3} 1.106 = 0.368

7 0
3 years ago
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