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ehidna [41]
3 years ago
8

What is this math problem? -15a=17

Mathematics
1 answer:
Anestetic [448]3 years ago
5 0

Answer:

17 67899887753 o

haggis to get a chance to get a chance

up to you and I don't know how I turned

him to be read by signs or something like that but

Step-by-step explanation:

join us for a while ago and I don't know

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The quadratic equation 2x^2+4x-30=0 has two solutions. one of the solutions is x=3 the other solution is?
BartSMP [9]

Answer:

2x^2+4x-30=0

Step-by-step explanation:

2x^2+10x-6x-30=0

2x(x+5)-6(x+5)=0

(2x-6)(x+5)=0

2x-6=0        x+5=0

x=6/2          x=-5

x=3

6 0
3 years ago
Read 2 more answers
One spring day, Landon noted the time of day and the temperature, in degrees Fahrenheit. His findings are as follows: At 6 a.m.,
Verizon [17]

Answer:

Graph these points listed below

Step-by-step explanation:

Graph the following points:

(6 am, 50)

(9 am, 59)

(1 pm, 63)

(6 pm, 63)

(8 pm, 59)

(12 am midnight, 58)

6 0
2 years ago
The dogs at a shelter are housed in 4 different rooms. If there are 10 dogs in each room, how many dogs are housed at the shelte
frozen [14]
If there are 10 dogs in each room there are 40 dogs
6 0
3 years ago
If x^2 - 16 > 0, which of the following must be true?
zimovet [89]
The left hand side of the equation is a difference of two squares and may be factored out as follows,
                              (x - 4)(x + 4) > 0
They may be individually taken as,
                                     x - 4 > 0    ; x > 4
                                     x + 4 > 0   ; x < -4
Thus, the answer to this item is letter A. 
5 0
3 years ago
I need help with #1 of this problem. It has writings on it because I just looked up the answer because I’m confused but I want t
belka [17]

In the figure below

1) Using the theorem of similar triangles (ΔBXY and ΔBAC),

\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}

Where

\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}

Thus,

\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}

thus, BC = 7.5

2) BX = 9, BA = 15, BY = 15, YC = y

In the above diagram,

\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}

Thus, from the theorem of similar triangles,

\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}

solving for y, we have

\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}

thus, YC = 10.

4 0
1 year ago
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