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3 years ago

Can someone help me in this question?

Mathematics

1 answer:

brilliants [131]3 years ago

5 0

The answer is H, if you plug in 13 for the x on both sides of the equation, you have 39 is less than 31 which is not true.

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What is the factored form of the expression? d^2 = 36d + 324

Veseljchak [2.6K]

Answer:

$(x-43.45)(x+7.45)=0$

Step-by-step explanation:

We have the quadratic equation $d^{2}=36d+324$ i.e. $d^{2}-36d-324=0$

As, the roots of the quadratic equation $ax^{2}+bx+c=0$ are given by $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ .

So, from the given equation, we have,

a = 1, b = -36 , c = -324.

Substituting the values in $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$ , we get,

$x=\frac{36\pm \sqrt{(-36)^{2}-4\times 1\times (-324)}}{2\times 1}$

i.e. $x=\frac{36\pm \sqrt{1296+1296}}{2}$

i.e. $x=\frac{36\pm \sqrt{2592}}{2}$

i.e. $x=\frac{36\pm 50.9}{2}$

i.e. $x=\frac{36+50.9}{2}$ and $x=\frac{36-50.9}{2}$

i.e. $x=\frac{86.9}{2}$ and $x=\frac{-14.9}{2}$

i.e. x = 43.45 and x = -7.45

Thus, the roots of the equation are 43.45 and -7.45.

Hence, the factored form of the given expression will be $(x-43.45)(x+7.45)=0$

3 0

3 years ago

Daniel got 86% on his test and he solved 5 points extra credit problem. What would be his new grade on this test?

tensa zangetsu [6.8K]

86%=86/100
86+5=91
91/100=91%

7 0

3 years ago

Please answer question three and four if you can :) Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0

3 years ago

HEEEEEEEEEEEEEEEEEEEEEELP 10 POINTS math vocab 2 questions

sergij07 [2.7K]

Answer:

Number 1: Evaluate (A)

Number 2: Operator (B)

6 0

3 years ago

Identify the like terms 3x,4y,4z,5x A:4y,5x B:4y,4z C:3x,5x D:3x,4z

fomenos

Answer:

Step-by-step explanation:

5 0

3 years ago