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Tems11 [23]
2 years ago
6

Plz help me If u do u are smart and gain 10 pts

Mathematics
1 answer:
Neporo4naja [7]2 years ago
4 0

Answer:

White:1.2kg

green:2kg

black:2.8

Step-by-step explanation:

Total ratio = 3+5+7=15

white :

=  \frac{3}{15}  \times 6

= 1.2

green:

\frac{5}{15}  \times 6

= 2

black:

\frac{7}{5}  \times 6

= 2.8

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Please help me with this!!!
RSB [31]
The answer that you have is most likely right, since the grouping did not change.
4 0
3 years ago
16 more than s is at most −80.
polet [3.4K]

Answer:

s + 16 ≤ −80

Step-by-step explanation:

Analyze each part in the word equation:

"16 more than s"     s increased by 16; s + 16

"is at most"           is equal to or less than; ≤

"−80"               -80

Put the equation back together:

s + 16 ≤ −80

4 0
4 years ago
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
When dividing the polynomial 3x3 - 5x2 - 4x + 4 by (x - 2), the remainder is 0, making (x - 2) a factor.
Ne4ueva [31]
(3x^3 - 5x^2 -4x + 4) / (x-2) = (3x-2)(x+1) when x ≠2
3 0
3 years ago
For f(x) = -4x -9 evaluate for f(6.5)​
kolezko [41]

Answer:

f=−35

Step-by-step explanation:

<em>1.Simplify  4×6.5 to 26</em>

f=-26-9

<em>2. Simplify  -26-9 to  -35</em>

f=-35

7 0
3 years ago
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