Answer:
The product of (x-3) and (x+7) is x² + 4x - 21
Step-by-step explanation:
The product of (x-3) and (x+7) is x² + 4x - 21
The product of (x - 3) and (x + 7) can be found as follows:
(x - 3)(x + 7)
open the bracket by multiplying
Therefore,
x(x) +x(7) - 3(x) - 3(7)
x² + 7x - 3x - 21
combine like terms
x² + 4x - 21
Based on the given polynomial, the degree of the polynomial can be calculated to be 1.
<h3>what is the degree of the polynomial?</h3>
the degree of the polynomial is defined as the highest exponential degree or power in a polynomial.
from the above, we see that the highest power is 1 from 8x¹.
the degree of the polynomial is therefore 1.
find out more on the degree of the polynomial at brainly.com/question/2263735.
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11.k>-3(4r+3)/4r-5
12.k<(2x+2)/x+3
(p + q)⁵
(p + q)(p + q)(p + q)(p + q)(p + q)
{[p(p + q) + q(p + q)][p(p + q) + q(p + q)](p + q)}
{[p(p) + p(q) + q(p) + q(q)][p(p) + p(q) + q(p) + q(q)](p + q)}
(p² + pq + pq + q²)(p² + pq + pq + q²)(p + q)
(p² + 2pq + q²)(p² + 2pq + q²)(p + q)
{[p²(p² + 2pq + q²) + 2pq(p² + 2pq + q²) + q²(p² + 2pq + q²)](p + q)}
{[p²(p²) + p²(2pq) + p²(q²) + 2pq(p²) + 2pq(2pq) + 2pq(q²) + q²(p²) + q²(2pq) + q²(q²)](p + q)}
(p⁴ + 2p³q + p²q² + 2p³q + 4p²q² + 2pq³ + p²q² + 2pq³ + q⁴)(p + q)
(p⁴ + 2p³q + 2p³q + p²q² + 4p²q² + p²q² + 2pq³ + 2pq³ + q⁴)(p + q)
(p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴)(p + q)
p⁴(p + q) + 4p³q(p + q) + 6p²q²(p + q) + 4pq³(p + q) + q⁴(p + q)
p⁴(p)+ p⁴(q) + 4p³q(p) + 4p³q(q) + 6p²q²(p) + 6p²q²(q) + 4pq³(p) + 4pq³(q) + q⁴(p) + q⁴(q)
p⁵ + p⁴q + 4p⁴q + 4p³q² + 6p³q² + 6p²q³ + 4p²q³ + 4pq⁴ + pq⁴ + q⁵
p⁵ + 5p⁴q + 10p³q² + 10p²q³ + 5pq⁴ + q⁵
So our equation to find this can be represented by 7x + 10 = 45.
(x representing miles run)
Just solve for x!
Subtract the 10 from both sides, you've got
7x = 35
And now to isolate the x, we divide both sides by 7.
Now we're left with just
x = 5!
We can check this by substituting 5 as x in our equation.
7x x 5 + 10 = 45
45 = 45
It's right!
So the runner ran 5 miles total.
Hope this helps!
If I skimmed over this too much, let me know and I'll try to explain the best that I can.
