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Arlecino [84]
2 years ago
14

If Alan read 204 pages in6 hours, how many pages would you expect him to read in 11 hours?

Mathematics
1 answer:
Amanda [17]2 years ago
4 0

Answer:

374 Pages

Step-by-step explanation:

First find the unit rate. 204/6=34 pages per hour.

34*11=374

*Please mark the brainliest*

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Someone that can help me out please
MAXImum [283]

Answer: I got 32

Step-by-step explanation:

I got 32

8 0
3 years ago
Solve for x<br><br> Please help me
Sedbober [7]

9514 1404 393

Answer:

  x = 8.2

Step-by-step explanation:

As indicated on the supplied diagram, the measure of arc GH is ...

  GH = 360° -83° -115° = 162°

The inscribed angle at F is half this measure, so we have ...

  (10x -1)° = 162°/2

  10x = 82 . . . . . . . . add 1 and simplify

  x = 8.2 . . . . . . . . . divide by 10

4 0
3 years ago
Please please help me with geometry
telo118 [61]

These two angles are alternate interior, meaning that they are the same.

Therefore, 11x-10 = 2x+17

Solve to get x = 3; the two angles both equal to 23 degrees

and since EGF and DGE together form 90, DGE = 90-23 = 67

5 0
2 years ago
Read 2 more answers
a rectangular storage container without a lid is to have a volume of 10 m3. the length of its base is twice the width. material
victus00 [196]

245.31 (dollars) is the cost (in dollars) of materials for the least expensive such container.

<h3>What is maxima and minima ? </h3>

Calculus maxima and minima are found  using the concept of derivatives. Knowing that the derivative concept gives information about the slope/slope of a function, we find the point where the slope is zero. These points are called inflection points/stationary points. These are the points associated with the maximum or minimum (local) values ​​of the function.

Knowledge of maxima and minima is essential for our everyday problems. In addition, this article also explains how to find the absolute maximum and minimum values.

Solvable maximum and minimum arithmetic problems  are discussed in this article.

<h3>Calculation</h3>

Suppose the width is x (m), length of the base is 2x (m), the base area is 2x^2 (m^2).

Since the volume is 10 (m^3), the height has to be 10/2x^2 (m) = 5/x^2.

The cost of making such container is

cost of base: 2x^2*15 = 30x^2

cost of sides: (2*2x*5/x^2 + 2*x*5/x^2)*9 = 270/x

The overall cost is hence the sum of the base and the sides: f(x) = 30x^2 + 270/x

The get the minimum,

df(x)/dx = 30*(2x - 9/x^2) = 0

so x = (9/2)^(1/3) = 1.651 (m)

f(x) = 245.31 (dollars)

learn more about maxima and minima here :

brainly.com/question/17467131

#SPJ4

8 0
1 year ago
Plzzzzzzz give me notes of dna replication​
miskamm [114]
- A DNA strand can act as a template for synthesis of a new nucleic acid strand in which each base forms a hydrogen-bonded pair with one on the template strand (G with C, A with T, or A with U for RNA molecules). The new sequence is thus complementary to the template strand. The copying of DNA molecules to produce more DNA is known as DNA Replication.

-DNA replication takes place at a Y-shaped structure called a replication fork. A self-correcting DNA polymerase enzyme catalyzes nucleotide polymerization in a 5ʹ-to-3ʹ direction, copying a DNA template strand with remarkable fidelity. Since the two strands of a DNA double helix are antiparallel, this 5ʹ-to-3ʹ DNA synthesis can take place continuously on only one of the strands at a replication fork (the leading strand).

-On the lagging strand, short DNA fragments must be made by a “backstitching” process. Because the self-correcting DNA polymerase cannot start a new chain, these lagging-strand DNA fragments are primed by short RNA primer molecules that are subsequently erased and replaced with DNA.

-DNA replication requires the cooperation of many proteins. These include:

*DNA polymerase and DNA primase to catalyze nucleoside triphosphate polymerization;
*DNA helicases and single-strand DNA-binding (SSB) proteins to help in opening up the DNA helix so that it can be copied;
*DNA ligase and an enzyme that degrades *RNA primers to seal together the discontinuously synthesized laggingstrand DNA fragments;
*DNA topoisomerases to help to relieve helical winding and DNA tangling problems. *Many of these proteins associate with each
other at a replication fork to form a highly efficient “replication machine,” through which the activities and spatial movements of the individual components are coordinated.

Major steps involved in DNA replication are as follows:

*Each strand in a parental duplex DNA acts as a template for synthesis of a daughter strand and remains basepaired to the new strand, forming a daughter duplex (semiconservative mechanism).
*New strands are formed in the 5′ to 3′ direction.
*Replication begins at a sequence called an origin.
*Each eukaryotic chromosomal DNA molecule contains multiple replication origins.
*DNA polymerases, unlike RNA polymerases, cannot unwind the strands of duplex DNA and cannot initiate synthesis of new strands complementary to the template strands.
*Helicases use energy from ATP hydrolysis to separate the parental (template) DNA strands.
*Primase synthesizes a short RNA primer, which remains base-paired to the template DNA.
*This initially is extended at the 3′ end by DNA polymerase α (Pol α), resulting in a short (5′ )RNA- (3′)DNA daughter strand.
*Most of the DNA in eukaryotic cells is synthesized by Pol ẟ, which takes over from Pol α and continues elongation of the daughter strand in the 5′ to 3’direction.
*Pol ẟ remains stably associated with the template by binding to Rfc protein, which in turn binds to PCNA, a trimeric protein that
encircles the daughter duplex DNA.
*DNA replication generally occurs by a bidirectional mechanism in which two replication forks form at an origin and move in opposite directions, with both template strands being copied at each fork.
*Synthesis of eukaryotic DNA in vivo is regulated by controlling the activity of the MCM helicases that initiate DNA replication at multiple origins spaced along chromosomal DNA.
*At a replication fork, one daughter strand (the leading strand) is elongated continuously.
*The other daughter strand (the lagging strand) is formed as a series of discontinuous Okazaki fragments from primers synthesized every few hundred nucleotides.
*The ribonucleotides at the 5′ end of each Okazaki fragment are removed and replaced by elongation of the 3′ end of the next Okazaki fragment.
*Finally, adjacent Okazaki fragments are joined by DNA ligase.
4 0
3 years ago
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