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1 year ago

Can someone explain this pls

Mathematics

1 answer:

Ludmilka [50]1 year ago

4 0

Answer:

Step-by-step explanation:

( g ° f )(x) = ( x + 7 )² = x² + 14x + 49

20 + 3(x² + 14x + 49) = - 34

20 + 3x² + 42x + 147 = - 34

3x² + 42x + 201 = 0 ⇔ x² + 14x + 67 = 0

D = 196 - 268 = - 72 < 0

There is no any roots in set of real numbers.

The values of x in set of complex numbers are:

$x_{12}$ = $\frac{- 14 }{2}$ ± $\frac{6\sqrt{2}i }{2}$

$x_{12}$ = - 7 ± 3√2 i

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Can someone solve this

Natalka [10]

Option D:

ΔCAN ≅ ΔWNA by SAS congruence rule.

Solution:

Given data:

m∠CNA = m∠WAN and CN = WA

To prove that ΔCAN ≅ ΔWNA:

In ΔCAN and ΔWNA,

CN = WA (given side)

∠CNA = ∠WAN (given angle)

NA = NA (reflexive side)

Therefore, ΔCAN ≅ ΔWNA by SAS congruence rule.

Hence option D is the correct answer.

4 0

2 years ago

A) 121 <br> B) 102<br> C) 141 <br> D) 115

patriot [66]

Answer:

121

Step-by-step explanation:

cause if its on the oppsite side its going to be the same number

8 0

2 years ago

Each day, a salesperson earns $25 plus 5% of their total sales for the day. The salesperson's goal is to earn more than $50 on T

uysha [10]

They need to sell $500 of stuff

4 0

1 year ago

Vern has just opened a flower shop. Because of the startup costs, for the next month his net earnings for each bouquet are −$2.5

Basile [38]

Answer:

-$200

Step-by-step explanation:

Given that:

Net earning per bouquet for the month= - $2.50

On a particular day of the month, 20 bouquet was sold ;

Net earning for the day:

Net earning per bouquet * number of bouquet

-$2.50 * 80

= $200

Hence, net earning for Tuesday = - $200

6 0

2 years ago

The graph of a linear function passes through the points (2, 4) and (8, 10).

Keith_Richards [23]

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 2}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ 8}}\quad ,&{{ 10}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{10-4}{8-2}$

$\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=2\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}$

8 0

2 years ago