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valkas [14]
3 years ago
9

PLEASE HELP!!

Mathematics
2 answers:
hjlf3 years ago
8 0
Hey idk if right but c I think
Ede4ka [16]3 years ago
3 0
I think the answer is -3
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Can someone solve and explain this question plz​
oee [108]

firstly let's convert the mixed fraction to improper fraction, then hmmm let's see we have two denominators, 5 and 3, and their LCD will simply be 15, so we'll multiply both sides by that LCD to do away with the denominators, let's proceed,

\bf \stackrel{mixed}{2\frac{1}{3}}\implies \cfrac{2\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{7}{3}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{z}{5}-4=\cfrac{7}{3}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{15}}{15\left( \cfrac{z}{5}-4 \right)=15\left( \cfrac{7}{3} \right)}\implies 3z-60=35 \\\\\\ 3z=95\implies z=\cfrac{95}{3}\implies z = 31\frac{2}{3}

4 0
3 years ago
Help? Please. Thanks???
Naya [18.7K]

The height is represented by d.

d = 10 feet

10 = -  16 {t}^{2}  - 7t + 61 \\ 0 =  - 16 {t}^{2}  - 7t + 51

Use quadratic formula:

t = 1.58, -2.02 (reject)

1.58 seconds

3 0
3 years ago
What is the length of the segment whose endpoints are A(-4, 3) and B (10, 6)?
nlexa [21]
Apply Pythagoras:

length = sqrt( (10--4)² + (6-3)² ) = sqrt(205)
7 0
3 years ago
Read 2 more answers
A jar contains 8 red marbles numbered 1 to 8 and 7 blue marbles numbered 1 to 7. A marble is drawn at random from the jar. Find
Shkiper50 [21]

Answer:

Therefore the probability that the marble is blue or even numbered is \frac{11}{15}

Step-by-step explanation:

Probability: The ratio of favorable outcomes to the total outcomes.

It is denoted by P.

Probability= \frac{\textrm{favorable outcomes}}{\textrm{Total outcomes}}

Given that a jar contains 8 red marbles and 7 blue marbles.

Total number of marbles = (8+7) = 15

Let A = Event of getting a blue marble

B= Event of getting of even marble.

Even number blue marbles are 2, 4,6

Even number red marbles are 2, 4,6,8

The number of even marbles are =(3+4)=7

The probability of getting a blue marble is P(A)

=\frac{\textrm{Total number of blue marbles}}{\textrm{Total number of blue marbles}}

=\frac{7}{15}

The probability of getting a even marble  is P(B)

=\frac{\textrm{The number of even number marbles}}{\textrm{Total number of marbles}}

=\frac{7}{15}

The probability of getting a even numbered blue marble P(A∩B)

=\frac{3}{16}

P(blue marble or even- numbered)

=P(A∪B)

=P(A)+P(B)-P(A∩B)

=\frac{7}{15} +\frac{7}{15}-\frac{3}{15}

=\frac{11}{15}

Therefore the probability that the marble is blue or even numbered is \frac{11}{15}

3 0
4 years ago
Expressed in the simplest form csc theta times tan theta times cos theta is equivalent to
mart [117]
<span>From the given information cosec theta times tan theta times cos theta, writing this triginometric expression as an equation. cosec theta x tan theta x cos theta cosec theta = 1/sin theta tan theta = sin theta / cos theta Calculating the equation after substituting (1/sin theta) x (sin theta / cos theta) x (cos theta) = 1</span>
4 0
3 years ago
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